Question 1:
In a car collision test, a car of mass 1 500 kg hits the wall with a speed of 15 m s–1. The car bounces back with a speed of 2.6 m s–1. If the collision time is 0.15 s, calculate the:
(i) impulse in the collision
(ii) impulsive force acting on the car
Answer:
(i)
$$ \begin{aligned} &\text { Impulse }(F t) \text { is a change of momentum }\\ &\begin{aligned} F t & =m v-m u \\ & =m(v-u) \\ & =1500[2.6-(-15)] \\ & =1500(17.6) \\ & =26400 \mathrm{kgms}^{-1} \\ \text { or } & =26400 \mathrm{Ns} \end{aligned} \end{aligned} $$
(ii)
$$ \text { Impulsive force is the rate of change of momentum } $$
$$ \begin{aligned} &\text { Daya impulse, } F\\ &\begin{aligned} & =\frac{m v-m u}{t} \\ & =\frac{26400}{0.15} \\ & =176000 \mathrm{~N} \end{aligned} \end{aligned} $$
In a car collision test, a car of mass 1 500 kg hits the wall with a speed of 15 m s–1. The car bounces back with a speed of 2.6 m s–1. If the collision time is 0.15 s, calculate the:
(i) impulse in the collision
(ii) impulsive force acting on the car
Answer:
(i)
$$ \begin{aligned} &\text { Impulse }(F t) \text { is a change of momentum }\\ &\begin{aligned} F t & =m v-m u \\ & =m(v-u) \\ & =1500[2.6-(-15)] \\ & =1500(17.6) \\ & =26400 \mathrm{kgms}^{-1} \\ \text { or } & =26400 \mathrm{Ns} \end{aligned} \end{aligned} $$
(ii)
$$ \text { Impulsive force is the rate of change of momentum } $$
$$ \begin{aligned} &\text { Daya impulse, } F\\ &\begin{aligned} & =\frac{m v-m u}{t} \\ & =\frac{26400}{0.15} \\ & =176000 \mathrm{~N} \end{aligned} \end{aligned} $$
Question 2:
A football player kicks a ball of mass 450 g with a force of 1 500 N. Th e contact time of his shoe with the ball is 0.008 s. What is the impulse on the ball? If contact time is increased to 0.013 s, what is the velocity of the ball?
Answer:
$$ \begin{aligned} m & =450 \mathrm{~g}=0.45 \mathrm{~kg} \\ F & =1500 \mathrm{~N} \\ t & =0.008 \mathrm{~s} \\ u & =0 \mathrm{~ms}^{-1} \end{aligned} $$
$$ \begin{aligned} \text { Impulse } & =F t \\ & =1500(0.008) \\ & =12 \mathrm{Ns} \end{aligned} $$
$$ \begin{aligned} t= & 0.013 \mathrm{~s} \\ F t & =m v-m u \\ 1500(0.013) & =0.45(v)-0 \\ 19.5 & =0.45 v \\ v & =\frac{19.5}{0.45} \\ & =43.33 \mathrm{~ms}^{-1} \end{aligned} $$
A football player kicks a ball of mass 450 g with a force of 1 500 N. Th e contact time of his shoe with the ball is 0.008 s. What is the impulse on the ball? If contact time is increased to 0.013 s, what is the velocity of the ball?
Answer:
$$ \begin{aligned} m & =450 \mathrm{~g}=0.45 \mathrm{~kg} \\ F & =1500 \mathrm{~N} \\ t & =0.008 \mathrm{~s} \\ u & =0 \mathrm{~ms}^{-1} \end{aligned} $$
$$ \begin{aligned} \text { Impulse } & =F t \\ & =1500(0.008) \\ & =12 \mathrm{Ns} \end{aligned} $$
$$ \begin{aligned} t= & 0.013 \mathrm{~s} \\ F t & =m v-m u \\ 1500(0.013) & =0.45(v)-0 \\ 19.5 & =0.45 v \\ v & =\frac{19.5}{0.45} \\ & =43.33 \mathrm{~ms}^{-1} \end{aligned} $$