FM1 §1.2 Conservation of momentum

Further mechanics 1

Table of contents

1. Principle of conservation of momentum
2. Examples
3. Edexcel FM1 Exercise 1B

1. Principle of conservation of momentum

We can solve problems involving collisions using the principle of conservation of momentum.

 

When two bodies collide, by Newton's 3rd law - the principle of action and reaction - each one exerts an equal and opposite force on the other. They are in contact for the same time, so they each exert an impulse on the other of equal magnitude but in opposite directions.

 

1. Let $ F_{12} $ be the force exerted by body 1 on body 2.
2. Let $ F_{21} $ be the force exerted by body 2 on body 1. We find: $$ F_{12} = - F_{21} $$
3. Then, the impulse exerted by body 1 on body 2 and that by body 2 on body 1 read $$ \begin{align} I_{12} &= F_{12} \Delta t = - F_{21} \Delta t = - I_{21} \end{align} $$

 

By the impulse-momentum principle, the changes in momentum of each body are equal but opposite in direction. Thus, these changes in momentum cancel each other, and the momentum of the whole system is unchanged. This is called the principle of conservation of momentum.

 

1. For body 1, the impulse-momentum principle gives $$ \begin{align} I_{21} = \Delta p_1 = m_1v_1 - m_1u_1 \end{align} $$
2. For body 2, the impulse-momentum principle gives $$ \begin{align} I_{12} = \Delta p_2 = m_2v_2 - m_2u_2 \end{align} $$
3. Since $ - I_{21} = I_{12}$, we find: $$ \begin{align} && - ( m_1v_1 - m_1u_1 ) &= m_2v_2 - m_2u_2 \\ \\ &\Rightarrow& m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ &\Leftrightarrow& \textrm{Total momentum before collision} &= \textrm{Total momentum after collision} \end{align} $$

 

When solving problems involving collisions:

1. draw a diagram showing the velocities before and after the collision with arrows
2. if appropriate, include the impulses on your diagram with arrows
3. choose a positive direction and apply the impulse-momentum principle and/or the principle of conservation of momentum.

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2. Examples

Example 1. A particle $P$ of mass 2 kg is moving with speed $ 3 \, \textrm{ms}^{-1} $ on a smooth horizontal plane. Particle $Q$ of mass 3 kg is at rest on the plane. Particle $P$ collides with particle $Q$ and after the collision $Q$ moves off with speed $ \frac73 \, \textrm{ms}^{-1} $. Find:

(a) the speed and direction of motion of $P$ after the collision

(b) the magnitude of the impulse received by $P$ in the collision.

 

Solution.

(a) By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & 2 \times 3 + 3 \times 0 &= 2 v_1 + 3 \times \frac73 \\ & \Rightarrow & v_1 &= \frac{6 - 7}{2} = - \frac12 \, \textrm{ms}^{-1} \end{align} $$ Since $v_1 < 0$, the direction of motion of $P$ is reversed by the collision and its speed is $0.5\,\textrm{ms}^{-1}$.

 

(b) By the impulse-momentum principle, $$ \begin{align} I &= \Delta p_1 \\ &= m_1 v_1 - m_1 u_1 \\ &= m_1 ( v_1 - u_1 ) \\ &= 2 \left( -\frac12 - 3 \right) \\ &= -7 \, \textrm{Ns} \end{align} $$ so the direction of the impulse received by $P$ is against its original motion and its magnitude is $7 \, \textrm{Ns}$.

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Example 2. Two particles $A$ and $B$ of masses 2 kg and 4 kg respectively are moving towards each other in opposite directions along the same straight line on a smooth horizontal surface. The particles collide. Before the collision the speeds of $A$ and $B$ are $ 3 \, \textrm{ms}^{-1} $ and $ 2 \, \textrm{ms}^{-1} $ respectively. After the collision the direction of motion of $A$ is reversed and its speed is $ 2 \, \textrm{ms}^{-1} $. Find:

(a) the speed and direction of motion of $B$ after the collision

(b) the magnitude of the impulse given by $A$ to $B$ in the collision.

 

Solution.

(a) By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & 2 \times 3 + 4 \times (-2) &= 2 \times (-2) + 4 \times v_2 \\ & \Rightarrow & v_2 &= \frac{6 - 8 + 4}{4} = \frac12 \, \textrm{ms}^{-1} \end{align} $$ Since $v_2 > 0$, the direction of motion of $B$ is reversed by the collision and its speed is $0.5\,\textrm{ms}^{-1}$.

 

(b) By the impulse-momentum principle, the impulse given by $A$ to $B$ is $$ \begin{align} I &= \Delta p_2 \\ &= m_2 v_2 - m_2 u_2 \\ &= m_2 ( v_2 - u_2 ) \\ &= 4 \left( \frac12 - (- 2) \right) \\ &= 10 \, \textrm{Ns} \end{align} $$ so the direction of the impulse received by $B$ is against its original motion and its magnitude is $10 \, \textrm{Ns}$.

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Example 3. Two particles $P$ and $Q$, of masses 8 kg and 2 kg respectively, are connected by a light inextensible string. The particles are at rest on a smooth horizontal plane with the string slack. Particle $P$ is projected directly away from $Q$ with speed $ 4 \, \textrm{ms}^{-1} $.

(a) Find the common speed of the particles after the string goes taut.

(b) Find the magnitude of the impulse transmitted through the string when it goes taut.

 

Solution.

(a) By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & 8 \times 4 + 2 \times 0 &= ( 8 + 2 ) v \\ & \Rightarrow & v &= \frac{32}{10} = \frac{16}{5} = 3.2 \, \textrm{ms}^{-1} \end{align} $$

 

(b) By the impulse-momentum principle, the impulse given by $P$ to $Q$ is $$ \begin{align} I_{PQ} &= \Delta p_2 \\ &= m_2 v_2 - m_2 u_2 \\ &= m_2 ( v_2 - u_2 ) \\ &= 2 \left( \frac{16}{5} - 0 \right) \\ &= \frac{32}{5} \\ &= 6.4 \, \textrm{Ns} \end{align} $$ Similarly, the impulse given by $Q$ to $P$ is $$ \begin{align} I_{QP} &= \Delta p_1 \\ &= m_1 v_1 - m_1 u_1 \\ &= 8 \left( \frac{16}{5} - 4 \right) \\ &= - \frac{32}{5} \\ &= - 6.4 \, \textrm{Ns} \end{align} $$ so the magnitude of the impulse is $6.4 \, \textrm{Ns}$.

 

Note: The impulse transmitted through the string is related to the derivative of acceleration, which is often called the 'jerk', i.e. $$ \begin{align} \textbf j = \frac{ \textrm d \textbf a }{ \textrm d t } = \frac{ \textrm d^2 \textbf v }{ \textrm d t^2 } = \frac{ \textrm d^3 \textbf x }{ \textrm d t^3 } \end{align} $$ and we also have higher derivatives:

$$ \begin{align} \textrm{Snap :}&& \textbf s &= \frac{ \textrm d \textbf j }{ \textrm d t } = \frac{ \textrm d^2 \textbf a }{ \textrm d t^2 } = \frac{ \textrm d^3 \textbf v }{ \textrm d t^3 } = \frac{ \textrm d^4 \textbf x }{ \textrm d t^4 } \\ \textrm{Crackle :}&& \textbf c &= \frac{ \textrm d \textbf s }{ \textrm d t } = \frac{ \textrm d^2 \textbf j }{ \textrm d t^2 } = \frac{ \textrm d^3 \textbf a }{ \textrm d t^3 } = \frac{ \textrm d^4 \textbf v }{ \textrm d t^4 } = \frac{ \textrm d^5 \textbf x }{ \textrm d t^5 } \\ \textrm{Pop :}&& \textbf p &= \frac{ \textrm d \textbf c }{ \textrm d t } = \frac{ \textrm d^2 \textbf s }{ \textrm d t^2 } = \frac{ \textrm d^3 \textbf j }{ \textrm d t^3 } = \frac{ \textrm d^4 \textbf a }{ \textrm d t^4 } = \frac{ \textrm d^5 \textbf v }{ \textrm d t^5 } = \frac{ \textrm d^6 \textbf x }{ \textrm d t^6 } \\ \end{align} $$

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Example 4. Two particles $A$ and $B$ of masses 2 kg and 4 kg respectively are moving towards each other in opposite directions along the same straight line on a smooth horizontal surface. The particles collide. Before the collision the speeds of $A$ and $B$ are $ 3 \, \textrm{ms}^{-1} $ and $ 2 \, \textrm{ms}^{-1} $ respectively. Given that the magnitude of the impulse due to the collision is 7 Ns, find:

(a) the velocity of $A$ after the collision

(b) the velocity of $B$ after the collision.

 

Solution.

(a) By the impulse-momentum principle, the impulse given by $B$ to $A$ is (it is in the negative direction) $$ \begin{align} && I_{BA} &= \Delta p_1 \\ \\ &\Rightarrow& -7 &= m_1 v_1 - m_1 u_1 \\ &&&= m_1 ( v_1 - u_1 ) \\ &&&= 2 \left( v_1 - 3 \right) \\ \\ &\Rightarrow& v_1 &= \frac{-7+2 \times3 }{2} = -\frac{1}{2} = -0.5 \, \textrm{ms}^{-1} \end{align} $$ so the direction of motion of $A$ is reversed by the collision.

 

(b) By the impulse-momentum principle, the impulse given by $A$ to $B$ is (it is in the positive direction) $$ \begin{align} && I_{AB} &= \Delta p_2 \\ \\ &\Rightarrow& 7 &= m_2 v_2 - m_2 u_2 \\ &&&= m_2 ( v_2 - u_2 ) \\ &&&= 4 \big[ v_2 - (-2) \big] \\ \\ &\Rightarrow& v_2 &= \frac{7-4 \times2 }{4} = -\frac{1}{4} = -0.25 \, \textrm{ms}^{-1} \end{align} $$ so the direction of motion of $B$ is unchanged by the collision.

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4. Edexcel Exercise 1B

Question 1. A particle $P$ of mass 2 kg is moving on a smooth horizontal plane with speed $ 4 \, \textrm{ms}^{-1} $. It collides with a second particle $Q$ of mass 1 kg which is at rest. After the collision $P$ has speed $ 2 \, \textrm{ms}^{-1} $ and it continues to move in the same direction. Find the speed of $Q$ after the collision.

 

Solution.

By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & 2 \times 4 + 1 \times 0 &= 2 \times 2 + 1 \times v_2 \\ & \Rightarrow & v_2 &= 8 - 4 = 4 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

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Question 2. A railway truck of mass 25 tonnes moving at $ 4 \, \textrm{ms}^{-1} $ collides with a stationary truck of mass 20 tonnes. As a result of the collision the trucks couple together. Find the common speed of the trucks after the collision.

 

Solution.

By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 = (m_1+m_2) v \\ \\ & \Rightarrow & 25000 \times 4 + 20000 \times 0 &= ( 25000 + 20000 ) \times v \\ & \Rightarrow & v &= \frac{ 100000 }{ 45000 } = \frac{20}{9} \simeq 2.22 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

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Question 3. Particles $A$ and $B$ have masses 0.5 kg and 0.2 kg respectively. They are moving with speeds $ 5 \, \textrm{ms}^{-1} $ and $ 2 \, \textrm{ms}^{-1} $ respectively in the same direction along the same straight line on a smooth horizontal surface when they collide. After the collision $A$ continues to move in the same direction with speed $ 4 \, \textrm{ms}^{-1} $. Find the speed of $B$ after the collision.

 

Solution.

By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & 0.5 \times 5 + 0.2 \times 2 &= 0.5 \times 4 + 0.2 v_2 \\ & \Rightarrow & v_2 &= \frac{ 2.5 + 0.4 - 2 }{ 0.2 } = 4.5 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

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Question 4. A particle of mass 2 kg is moving on a smooth horizontal plane with speed $ 4 \, \textrm{ms}^{-1} $. It collides with a second particle of mass 1 kg which is at rest. After the collision the particles join together.

(a) Find the common speed of the particles after the collision.

(b) Find the magnitude of the impulse in the collision.

 

Solution.

(a) By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 = ( m_1 + m_2 ) v \\ \\ & \Rightarrow & 2 \times 4 + 1 \times 0 &= 3 v \\ & \Rightarrow & v &= \frac{ 8 }{ 3 } \simeq 2.67 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

 

(b) By the impulse-momentum principle, the impulse given by $A$ to $B$ is (it is in the positive direction) $$ \begin{align} I_{AB} &= \Delta p_2 \\ \\ &= m_2 v_2 - m_2 u_2 \\ &= m_2 ( v_2 - u_2 ) \\ &= 1 \left( \frac83 - 0 \right) \\ &= \frac83 \, \textrm{Ns} \qquad \checkmark \end{align} $$ Or, equivalently, the impulse given by $A$ to $B$ is (it is in the negative direction) $$ \begin{align} I_{BA} &= \Delta p_1 \\ \\ &= m_1 v_1 - m_1 u_1 \\ &= m_1 ( v_1 - u_1 ) \\ &= 2 \left( \frac83 - 4 \right) \\ &= - \frac83 \, \textrm{Ns} \qquad \checkmark \end{align} $$ So the magnitude of the impulse is $\frac83 \simeq 2.67 \, \textrm{Ns}$.

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Question 5. (E) Two particles $A$ and $B$ of masses 2 kg and 5 kg respectively are moving towards each other along the same straight line on a smooth horizontal surface. The particles collide. Before the collision the speeds of $A$ and $B$ are $ 6 \, \textrm{ms}^{-1} $ and $ 4 \, \textrm{ms}^{-1} $ respectively. After the collision the direction of motion of $A$ is reversed and its speed is $ 1.5 \, \textrm{ms}^{-1} $. Find:

(a) the speed and direction of $B$ after the collision [3 marks]

(b) the magnitude of the impulse given by $A$ to $B$ in the collision.[3 marks]

 

Solution.

(a) By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & 2 \times 6 + 5 \times (-4) &= 2 \times (-1.5) + 5 v_2 \\ & \Rightarrow & v_2 &= \frac{ 12 - 20 + 3 }{ 5 } = -1 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$ So, the speed of $B$ after the collision is $ 1 \, \textrm{ms}^{-1} $ and the direction is unchanged by the collision.

 

(b) By the impulse-momentum principle, the impulse given by $A$ to $B$ is (it is in the positive direction) $$ \begin{align} I_{AB} &= \Delta p_2 \\ &= m_2 v_2 - m_2 u_2 \\ &= m_2 ( v_2 - u_2 ) \\ &= 5 \big[ -1 - (-4) \big] \\ &= 15 \, \textrm{Ns} \qquad \checkmark \end{align} $$

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Question 6. (E) A particle $P$ of mass 150 g is at rest on a smooth horizontal plane. A second particle $Q$ of mass 100 g is projected along the plane with speed $ u \, \textrm{ms}^{-1} $ and collides directly with $P$. On impact the particles join together and move on with speed $ 4 \, \textrm{ms}^{-1} $. Find the value of $u$. [4 marks]

 

Solution.

By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 = ( m_1 + m_2 ) v \\ \\ & \Rightarrow & 0.1 u + 0.15 \times 0 &= ( 0.1 + 0.15 ) \times 4 \\ & \Rightarrow & u &= \frac{ 0.25 \times 4 }{ 0.1 } = 10 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

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Question 7. (E/P) A particle $A$ of mass $4m$ is moving along a smooth horizontal surface with speed $2u$. It collides with another particle $B$ of mass $3m$ which is moving with the same speed along the same straight line but in the opposite direction. Given that $A$ is brought to rest by the collision, find:

(a) the velocity of $B$ after the collision [3 marks]

(b) the magnitude of the impulse given by $A$ to $B$ in the collision. [3 marks]

 

Solution.

(a) By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & 4m \times 2u + 3m \times (-2u) &= 4m \times 0 + 3m v_2 \\ & \Rightarrow & v_2 &= \frac{ 8mu - 6mu }{ 3m } = \frac23 u \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$ so the dirction of motion of $B$ is reversed by the collision.

 

(b) By the impulse-momentum principle, the impulse given by $A$ to $B$ is (it is in the positive direction) $$ \begin{align} I_{AB} &= \Delta p_2 \\ &= m_2 v_2 - m_2 u_2 \\ &= m_2 ( v_2 - u_2 ) \\ &= 3m \left[ \frac23 u - (-2u) \right] \\ &= 8mu \qquad \checkmark \end{align} $$

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Question 8. (E/P) An explosive charge of mass 150 g is designed to split into two parts, one with mass 100 g and the other with mass 50 g. When the charge is moving at $ 4 \, \textrm{ms}^{-1} $ it splits and the larger part continues to move in the same direction whilst the smaller part moves in the opposite direction. Given that the speed of the larger part is twice the speed of the smaller part, find the speeds of each of the two parts. [3 marks]

 

Solution.

By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & ( 0.05 + 0.1 ) \times 4 &= 0.05 \times (-v) + 0.1 \times 2v \\ & \Rightarrow & v &= \frac{ 0.15 \times 4 }{ 0.15 } = 4 \, \textrm{ms}^{-1} \end{align} $$ so the speed of the smaller part is $4 \, \textrm{ms}^{-1}$ and the speed of the larger part is $8 \, \textrm{ms}^{-1}$.

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Question 9. (E/P) Two particles $P$ and $Q$ of masses $m$ and $km$ respectively are moving towards each other in opposite directions along the same straight line on a smooth horizontal surface. The particles collide. Before the collision the speeds of $P$ and $Q$ are $3u$ and $u$ respectively. After the collision the direction of motion of both particles is reversed and the speed of each particle is halved.

(a) Find the value of $k$. [4 marks]

(b) Find, in terms of $m$ and $u$, the magnitude of the impulse given by $P$ to $Q$ in the collision. [3 marks]

 

Solution.

(a) By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & m \times 3u + km \times (-u) &= m \times \left( - \frac32 u \right) + km \times \frac12 u \\ & \Rightarrow & 3 - k &= - \frac32 + \frac12 k \\ & \Rightarrow & \left( 1 + \frac12 \right) k &= 3 + \frac32 \\ & \Rightarrow & k &= \frac{ 3 + \frac32 }{ \frac32 } = 3 \qquad \checkmark \end{align} $$

 

(b) By the impulse-momentum principle, the impulse given by $P$ to $Q$ is (it is in the positive direction) $$ \begin{align} I_{PQ} &= \Delta p_2 \\ &= m_2 v_2 - m_2 u_2 \\ &= m_2 ( v_2 - u_2 ) \\ &= km \left[ \frac12 u - (-u) \right] \\ &= 3m \times \frac32 u \\ &= \frac{9mu}{2} \qquad \checkmark \end{align} $$

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Question 10. (E/P) Two particles $A$ and $B$ of masses 4 kg and 2 kg respectively are connected by a light inextensible string. The particles are at rest on a smooth horizontal plane with the string slack. Particle $A$ is projected directly away from $B$ with speed $ u \, \textrm{ms}^{-1} $. When the string goes taut the impulse transmitted through the string has magnitude $6 \, \textrm{Ns}$. Find:

(a) the common speed of the particles just after the string goes taut [4 marks]

(b) the value of $u$. [3 marks]

 

Solution.

(a) By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 = (m_1 + m_2 ) v \\ \\ & \Rightarrow & 0 + 4 \times u &= 6v \\ & \Rightarrow & u &= \frac32 v \qquad (*) \end{align} $$

By the impulse-momentum principle, the impulse given by $A$ to $B$ is (it is in the positive direction) $$ \begin{align} && I_{AB} &= \Delta p_1 \\ \\ &\Rightarrow& 6 &= m_1 v_1 - m_1 u_1 \\ &&&= m_1 ( v_1 - u_1 ) \\ &&&= 2 v \\ \\ &\Rightarrow& v &= 3 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

 

(b) From the momentum conservation equation (*), $$ \begin{align} u = \frac{3}{2}v = \frac{9}{2} \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

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Question 11. (E/P) Two particles $P$ and $Q$ of masses 3 kg and 2 kg respectively are moving along the same straight line on a smooth horizontal surface. The particles collide. After the collision both the particles are moving in the same direction, the speed of $P$ is $ 1 \, \textrm{ms}^{-1} $ and the speed of $Q$ is $ 1.5 \, \textrm{ms}^{-1} $. The magnitude of the impulse of $P$ on $Q$ is $9 \, \textrm{Ns}$. Find:

(a) the speed and direction of $P$ before the collision [3 marks]

(b) the speed and direction of $Q$ before the collision. [3 marks]

 

Solution.

(a) By the impulse-momentum principle, we consider the impulse given by $Q$ to $P$. It is only given that they move in the same direction after the collision. However, by looking at their speeds, since $Q$ moves faster than $P$ in the same direction, we may assume that the direction of motion of $Q$ after the collision is in the positive ($\overrightarrow{PQ}$) direction. If on the other hand $P$ moved faster than $Q$ in the same direction after the collision, it means that their velocities are in the negative ($\overrightarrow{QP}$) direction. $$ \begin{align} && I_{QP} &= \Delta p_1 \\ \\ &\Rightarrow& -9 &= m_1 v_1 - m_1 u_1 \\ &&&= m_1 ( v_1 - u_1 ) \\ &&&= 3 ( 1 - u_1 ) \\ \\ &\Rightarrow& u_1 &= 4 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$ so initially $P$ was moving towards $Q$ (i.e. in the positive direction) at $ 4 \, \textrm{ms}^{-1} $.

 

(b) By the impulse-momentum principle, we consider the impulse given by $P$ to $Q$. $$ \begin{align} && I_{PQ} &= \Delta p_2 \\ \\ &\Rightarrow& 9 &= m_2 v_2 - m_2 u_2 \\ &&&= m_2 ( v_2 - u_2 ) \\ &&&= 2 ( 1.5 - u_2 ) \\ \\ &\Rightarrow& u_2 &= - 3 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$ so initially $Q$ was moving towards $P$ at $ 3 \, \textrm{ms}^{-1} $.

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Question 12. (E/P) Two particles $A$ and $B$ are moving in the same direction along the same straight line on a smooth horizontal surface. The particles collide. Before the collision the speed of $B$ is $ 1.5 \, \textrm{ms}^{-1} $. After the collision the direction of motion of both particles is unchanged, the speed of $A$ is $ 2.5 \, \textrm{ms}^{-1} $ and the speed of $B$ is $ 3 \, \textrm{ms}^{-1} $. Given that the mass of $A$ is three times the mass of $B$,

(a) find the speed $A$ before the collision. [4 marks]

 

Given that the magnitude of the impulse on $A$ in the collision is $3 \, \textrm{Ns}$,

(b) find the mass of $A$. [3 marks]

 

Solution.

(a) Let the initial speed of $A$ be $u$ and the mass of $B$ be $m$. By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & 3m \times u + m \times 1.5 &= 3m \times 2.5 + m \times 3 \\ & \Rightarrow & 3u + 1.5 &= 7.5 + 3 \\ & \Rightarrow & u &= \frac{ 7.5 + 3 - 1.5 }{ 3 } = \frac{9}{3} = 3 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

 

(b) By the impulse-momentum principle, the impulse given by $B$ to $A$ is (it is in the negative direction) $$ \begin{align} && I_{BA} &= \Delta p_1 \\ \\ &\Rightarrow& -3 &= m_1 v_1 - m_1 u_1 \\ && &= m_1 ( v_1 - u_1 ) \\ && &= 3m (2.5 -3) \\ && &= -0.5 (3m) \\ \\ &\Rightarrow& m_A &= 3m = 6 \, \textrm{kg} \qquad \checkmark \end{align} $$ Or, equivalently, the impulse given by $B$ to $A$ is (it is in the positive direction) $$ \begin{align} && I_{AB} &= \Delta p_2 \\ \\ &\Rightarrow& 3 &= m_2 v_2 - m_2 u_2 \\ &&&= m_2 ( v_2 - u_2 ) \\ &&&= m (3 - 1.5) \\ &&&= 1.5m \\ \\ &\Rightarrow& m_B &= m = 2 \, \textrm{kg} \\ && m_A &= 3m = 6 \, \textrm{kg} \qquad \checkmark \end{align} $$

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Challenge. Particle $P$ has mass $3m$ kg and particle $Q$ has mass $m$ kg. The particles are moving in opposite directions along the same straight line on a smooth horizontal plane when they collide directly. Immediately before the collision, the speed of $P$ is $ u_1 \, \textrm{ms}^{-1} $ and the speed of $Q$ is $ u_2 \, \textrm{ms}^{-1} $. In the collision, the direction of motion of $P$ is unchanged and the direction of $Q$ is reversed. Immediately after the collision, the speed of $P$ is $ \frac14 u_1 \, \textrm{ms}^{-1} $ and the speed of $Q$ is $ \frac12 u_2 \, \textrm{ms}^{-1} $. Show that $ u_1 = \frac23 u_2 $.

 

Solution.

By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & 3m \times u_1 + m \times(-u_2) &= 3m \times \frac14 u_1 + m \times \frac12 u_2 \\ & \Rightarrow & 3u_1 - u_2 &= \frac34 u_1 + \frac12 u_2 \\ & \Rightarrow & \left( 3 - \frac34 \right) u_1 &= \left( 1 + \frac12 \right) u_2 \\ & \Rightarrow & \frac{ 9 }{ 4 } u_1 &= \frac{ 3 }{ 2 } u_2 \\ & \Rightarrow & u_1 &= \frac{ 2 }{ 3 } u_2 \qquad \checkmark \end{align} $$

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Investigation. (See also FM1 §4.3 Loss of kinetic energy) A bullet is fired horizontally from a rifle. The rifle has mass $M$ and the bullet has mass $m$. The initial speed of the bullet is $v$ and the initial speed with which the rifle recoils is $V$. Consider:

(i) momentum

(ii) kinetic energy

(iii) impulse

(iv) force

 

Solution.

(i) By the principle of conservation of momentum, $$ \begin{align} && m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \\ & \Rightarrow & 0 &= m v + M V \\ & \Rightarrow & V &= - \frac{ m }{ M }v \qquad \checkmark \end{align} $$

 

(ii) The kinetic energy reads $$ \begin{align} \textrm{Bullet :} && \textrm{KE}_{ \textrm{bullet} } &= \frac12 mv^2 \\ \textrm{Rifle :} && \textrm{KE}_{ \textrm{rifle} } &= \frac12 MV^2 \\ \\ \Rightarrow && \frac{ \textrm{KE}_{ \textrm{bullet} } }{ \textrm{KE}_{ \textrm{rifle} } } &= \frac{ \frac12 mv^2 }{ \frac12 MV^2 } = \frac{ mv^2 }{ M \left( - \frac{ m }{ M }v \right)^2 } = \frac{ M }{ m } \end{align} $$

 

(iii) By the impulse-momentum principle, the impulse given to the bullet is $$ \begin{align} I &= \Delta p = mv = - MV \end{align} $$

 

(iv) The force is the rate of change of momentum with respect to time, i.e. $$ \begin{align} F = \frac{ \Delta p }{ \Delta t } \end{align} $$ where $\Delta t$ denotes the duration over which the impulse is acted upon the bullet/rifle.

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A related question to investigation. A machine gun fires 35.0 g bullets at a speed of $ 750.0 \, \textrm{ms}^{-1} $. If the gun can fire 200 bullets/min, what is the average force the shooter must exert to keep the gun from moving?

 

Solution.

(i) For each bullet, the change in momentum is $$ \begin{align} \Delta p = mv = 0.035 \times 750 = 26.25 \, \textrm{kg ms}^{-1} \end{align} $$ (ii) Since the gun fires 200 bullets in 60 seconds, we may assume that each bullet gains momentum in: $$ \begin{align} \Delta t = \frac{ 60 \, \textrm{s} }{ 200 } = 0.3 \, \textrm{s} \end{align} $$ (iii) The force reads $$ \begin{align} F = \frac{ \Delta p }{ \Delta t } = \frac{ 26.25 \, \textrm{kg ms}^{-1} }{ 0.3 \, \textrm{s} } = 87.5 \, \textrm{N} \qquad \checkmark \end{align} $$