FM1 §1.1 Momentum in one direction

Further mechanics 1

Table of contents

1. Momentum = Linear momentum
2. Units
3. First example
4. Impulse
5. The impulse-momentum principle
6. More examples
7. Exercise 1A

1. Momentum = Linear momentum

The momentum $p$ of a body of mass $m$ which is moving with velocity $v$ is: $$ \boxed{ \begin{align} \textrm{momentum} &= \textrm{mass} \times \textrm{velocity} \\ p &= mv \end{align} } $$ Mass is a scalar and velocity if a vector quantity, so momentum is a vector quantity: $$ \boxed{ \begin{align} \textbf p = m \textbf v \end{align} } $$

 

There is an analogous quantity called the angular momentum defined by $$ \textbf L = \textbf r \times \textbf p $$ When both momenta are in play, in order to avoid any confusion, the momentum defined above is often called the linear momentum. Unless this is the case, we shall just call it 'momentum'.

2. Units

If $m$ is measured in kilograms and $v$ is measured in $\textrm{ms}^{-1}$, then the units of momentum will be $$ [p]=[m][v]= \textrm{kg}\,\textrm{ms}^{-1}.$$ However, since $\textrm{kg}\,\textrm{ms}^{-2}$ are the units of force, the units of momentum may be written as newton seconds (Ns). The newton seconds are motivated by the impulse-momentum principle, see below.

3. First example

Example 1. Find the magnitude of the momentum of

(a) a cricket ball of mass 400 g moving at $18 \, \textrm{ms}^{-1}$

(b) a lorry of mass 5 tonnes moving at $0.3 \, \textrm{ms}^{-1}$.

 

Solution.

(a) $$ \begin{align} p &= mv \\ &= 0.4 \, \textrm{kg} \times 18 \, \textrm{ms}^{-1} \\ &= 7.2 \, \textrm{kg}\,\textrm{ms}^{-1} \end{align} $$ (b) $$ \begin{align} p &= mv \\ &= 5000 \, \textrm{kg} \times 0.3 \, \textrm{ms}^{-1} \\ &= 1500 \, \textrm{kg}\,\textrm{ms}^{-1} \end{align} $$

4. Impulse

If a constant force $F$ acts on a body for time $t$, the impulse is of the force is defined by $$ I = F \times t $$ Force is a vector quantity and time is a scalar, so impulse is a vector quantity.

 

If force $F$ is measured in newtons (N) and time $t$ is measured in seconds (s), then the units of impulse is $$ [I] = \textrm{Ns} $$

Aside. Non-constant forces:

For a non-constant force, i.e. it is a function of time, or equivalently it varies over time $$ \textbf F = \textbf F(t) $$ the impulse of the force is defined by the time integral: $$ \begin{align} \textbf I = \int \textbf F(t) \, \textrm dt \end{align} $$ In other words, in the $F-t$ graph, the impulse is given by the area under the curve. For $F(t)=\textrm{constant}$, we recover the original exprssion, $I=Ft$.

 

Examples of an impulse include a bat hitting a ball, a snooker ball hitting another ball or a jerk in a string when it suddenly goes tight. In all these cases the time for which the force acts is very small but the force is quite large and so the product of the two, which gives the impulse, is of reasonable size. However, there is no theoretical limit on the size of $t$.

5. The impulse-momentum principle

The impulse-momentum principle states that the impulse given to the body is the same as the change in momentum of the body.

 

For a constant force $F$ (hence a constant acceleration), we can derive this by using one of the SUVAT equations. Recall that the SUVAT equations show the relations between $(u,v,a,s,t)$ when the acceleration is constant. $$ \begin{align} \textrm{Gradient of $v-t$ graph}:\qquad v &= u + at \\ \textrm{Area of $v-t$ graph}:\qquad s &= \left( \frac{u+v}{2} \right) t \\ &= ut + \frac12 at^2 \\ &= vt - \frac12at^2 \\ 2as &= v^2 - u^2 \end{align} $$ From the first equation, $$ \begin{align} && a &= \frac{v-u}{t} \\ &\Rightarrow& F = ma &= \frac{mv-mu}{t} \\ &\Rightarrow& I = Ft &= mv - mu \end{align} $$

 

Aside. A derivation from Newton's 2nd law:

For a non-constant force, $$ \begin{align} && F(t) &= m \frac{ \textrm dv}{ \textrm dt} \\ &\Rightarrow& I = \int F(t) \, \textrm dt &= \int m \, \textrm dv = mv - mu \end{align} $$ and thus $$ \textrm{Impulse} = \textrm{Final momentum} - \textrm{Initial momentum} = \Delta p $$

 

Comment. The equation above shows that the time-integral of Newton's second law gives the impulse-momentum principle. Also, in FM1 §1.2 Conservation of momentum, we will learn that the impulse-momentum principle leads to the principle of conservation of linear momentum.

In addition, in FM1 §2.3 Conservation of mechanical energy and the work-energy principle, we will learn that the space-integral (or $x$-integral) of Newton's second law gives the work-energy principle that leads to the principle of conservation of mechanical energy.

 

1. The $t$-integral of Newton's 2nd law gives the impulse-momentum principle and conservation of linear momentum.
2. The $x$-integral of Newton's 2nd law gives the work-energy principle and conservation of mechanical energy.

6. More examples

Example 2. A body of mass 2 kg is initially at rest on a smooth horizontal plane. A horizontal force of magnitude 4.5 N acts on the body for 6 s. Find:

(a) the magnitude of the impulse given to the body by the force

(b) the final speed of the body.

 

Solution.

(a)

$$ \begin{align} I = Ft = 4.5 \, \textrm N \times 6 \, \textrm s = 27 \, \textrm{Ns} \qquad \checkmark \end{align} $$

 

(b) With $u=0\,\textrm{ms}^{-1}$,

$$ \begin{align} && I &= \Delta p = mv - mu \\ \\ &\Rightarrow& v &= \frac{I}{m} = \frac{27}{2} = 13.5 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

 

Example 3. A ball of mass 0.2 kg hits a fixed vertical wall at right angles with speed $ 3.5 \, \textrm{ms}^{-1} $. The ball rebounds with speed $ 2.5 \, \textrm{ms}^{-1} $. Find the magnitude of the impulse exerted on the wall by the ball.

 

Solution. For the ball, the total change of momentum reads (assuming that the final motion is in the positive direction) $$ \begin{align} I = \Delta p = mv - mu = 0.2 \Big[ 2.5 - (-3.5) \Big] = 1.2 \, \textrm{Ns} \end{align} $$ This impulse is given by the force exerted on the ball by the wall. By Newton's 3rd law, an equal and opposite force must have exerted on the wall by the ball for the same amount of time as this is the duration of their contact - we will revisit this when deriving the conservation of momentum in FM1 §1.2 Conservation of momentum. Thus, the magnitude of the impulse exerted on the wall by the ball is also $ 1.2 \, \textrm{Ns} $ in the direction of the ball's initial motion.

 

7. Exercise 1A

Question 1. A ball of mass 0.5 kg is at rest when it is struck by a bat and receives an impulse of 15 Ns. Find its speed immediately after it is struck.

 

Solution.

By the impulse-momentum principle, with $u=0 \, \textrm{ms}^{-1}$, $$ \begin{align} && I &= \Delta p = mv - mu \\ \\ &\Rightarrow& v &= \frac{I}{m} = \frac{15}{0.5} = 30 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

 

Question 2. A ball of mass 0.3 kg moving along a horizontal surface hits a fixed vertical wall at right angles with speed $ 3.5 \, \textrm{ms}^{-1} $. The ball rebounds at right angles to the wall. Given that the magnitude of the impulse exerted on the ball by the wall is 1.8 Ns, find the speed of the ball just after it rebounds.

 

Solution.

By the impulse-momentum principle, assuming that the final motion of the ball is in the positive direction, $$ \begin{align} && I &= \Delta p = mv - mu \\ &\Rightarrow& 1.8 &= 0.3 \Big[ v - (-3.5) \Big] \\ \\ &\Rightarrow& v &= \frac{1.8 - 0.3 \times 3.5 }{ 0.3 } = 2.5 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

 

Question 3. A toy car of mass 0.2 kg is pushed from rest along a smooth horizontal floor by a horizontal force of magnitude 0.4 N for 1.5 s. Find its speed at the end of the 1.5 s.

 

Solution.

By the impulse-momentum principle, with $ u = 0 \, \textrm{ms}^{-1} $ $$ \begin{align} && I &= \Delta p = mv - mu \\ \\ &\Rightarrow& v &= \frac{I}{v} = \frac{Ft}{v} = \frac{0.4 \times 1.5}{0.2} = 3 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$

 

Question 4. (E) A ball of mass 0.2 kg, moving along a horizontal surface, hits a fixed vertical wall at right angles. The ball rebounds at right angles to the wall with speed $ 3.5 \, \textrm{ms}^{-1} $. Given that the magnitude of the impulse exerted on the ball by the wall is 2 Ns, find the speed of the ball just before it hits the wall. [3 marks]

 

Solution.

By Newton's 3rd law, an equal and opposite impulse is exerted on the ball by the wall. By the impulse-momentum principle, assuming that the final motion of the ball is in the positive direction, $$ \begin{align} && I &= \Delta p = mv - mu \\ &\Rightarrow& 2 &= 0.2 \Big[ 3.5 - u \Big] \\ \\ &\Rightarrow& u &= \frac{ 0.2 \times 3.5 - 2 }{ 0.2 } = -6.5 \, \textrm{ms}^{-1} \qquad \checkmark \end{align} $$ where the minus sign indicates that the initial velocity is towards the wall.

 

Question 5. (E/P) A ball of mass 0.2 kg is dropped from a height of 2.5 m above horizontal ground. After hitting the ground it bounces to a height of 1.8 m above the ground. Find the magnitude of the impulse received by the ball from the ground. [4 marks]

 

Solution.

(i) From $h=2.5$ m to the ground: the ball falls under gravity, i.e. with a constant acceleration. From the SUVAT equations, $$ \begin{align} && v^2 &= u^2 + 2as \\ &&&= 0^2 + 2 \times 9.8 \times 2.5 \\ &&&= 49 \\ \\ &\Rightarrow& v &= 7 \, \textrm{ms}^{-1} \end{align} $$ (ii) From the ground to $h=1.8$ m: similarly, $$ \begin{align} && v^2 &= u^2 + 2as \\ &\Rightarrow& 0 &= u^2 + 2 \times (-9.8) \times 1.8 \\ &\Rightarrow& u^2 &= \frac{882}{25} = 35.28 \\ \\ &\Rightarrow& u &= \frac{21\sqrt{2}}{5} = 5.94 \, \textrm{ms}^{-1} \end{align} $$ (iii) Impulse: The ball hits the ground with $7 \, \textrm{ms}^{-1}$ and bounces back with $5.94 \, \textrm{ms}^{-1}$, $$ \begin{align} I = \Delta p &= mv - mu \\ &= 0.2 \left[ \frac{21\sqrt{2}}{5} - (-7) \right] \\ &= \frac{ 7 \left( 5 + 3 \sqrt{2} \right) }{25} \\ &= 2.59 \, \textrm{Ns} \quad \textrm{(3 s.f.)} \qquad \checkmark \end{align} $$