Jan Dangerfield, Stuart Haring and, Julian Gilbey Solutions for Exercise 3: EXERCISE 8B

At its launch a rocket has mass $2$million $\mathrm{kg}$. It accelerates from rest to $75000{ \mathrm{ms}}^{-1}$.

Work out the increase in the kinetic energy.

At its launch a rocket has mass $2$ million $\mathrm{kg}$. It accelerates from rest to $75000{ \mathrm{ms}}^{-1}$.

Why will the calculated value be too big?

Ball $A$, of mass $2 \mathrm{kg}$, is moving in a straight line at $5{ \mathrm{ms}}^{-1}$. Ball $B$, of mass $4 \mathrm{kg}$, is moving in the same straight line at $2{ \mathrm{ms}}^{-1}$. Ball $B$ is travelling directly towards ball $A$. The balls hit each other and after the impact each ball has reversed its direction of travel. The kinetic energy lost in the impact is $12.5 \mathrm{J}$.
Show that the speed of ball $A$ after the impact is $\frac{10}{3} \mathrm{m}{ \mathrm{s}}^{-1}$.

Ball $A$, of mass $2 \mathrm{kg}$, is moving in a straight line at $5{ \mathrm{ms}}^{-1}$. Ball $B$, of mass $4 \mathrm{kg}$, is moving in the same straight line at $2{ \mathrm{ms}}^{-1}$. Ball $B$ is travelling directly towards ball $A$. The balls hit each other and after the impact each ball has reversed its direction of travel. The kinetic energy lost in the impact is $12.5 \mathrm{J}$.

Find the speed of ball $B$ after the impact.

Two balls, $A$ and $B$, of equal mass, are travelling towards one another with velocities $u_A$ and $u_B$, respectively. The balls collide and their velocities after the impact are $-v_A$and $v_B$, respectively. The kinetic energy after the impact is the same as the kinetic energy before the impact (i.e. a 'perfectly elastic collision'). Explain why $v_A=u_B$ and $v_B=u_A$.

Balls $X,Y$ and $Z$ lie at rest on a smooth horizontal surface, with $Y$ between $X$ and $Z$. Balls $X$ and $Z$ each have mass $2 \mathrm{kg}$ and ball $Y$ has mass $1 \mathrm{kg}$. Ball $X$ is given a velocity of $1 \mathrm{m}{ \mathrm{s}}^{-1}$ towards ball $Y$, Balls $X$ and $Y$ collide. After this collision the speed of ball $X$ is $0.4{ \mathrm{ms}}^{-1}$ in its original direction.

Work out the loss in kinetic energy in this impact.

Ball $Y$ goes on to collide with ball $Z$. After this collision the speed of ball $Y$ is $0.4 \mathrm{m}{ \mathrm{s}}^{-1}$ in the direction towards ball $X$.

Balls $X,Y$ and $Z$ lie at rest on a smooth horizontal surface, with $Y$ between $X$ and $Z$. Balls $X$ and $Z$ each have mass $2 \mathrm{kg}$ and ball $Y$ has mass $1 \mathrm{kg}$. Ball $X$ is given a velocity of $1 \mathrm{m}{ \mathrm{s}}^{-1}$ towards ball $Y$, Balls $X$ and $Y$ collide. After this collision the speed of ball $X$ is $0.4{ \mathrm{ms}}^{-1}$ in its original direction.

b Work out the loss in kinetic energy in this impact. Finally, ball $Y$ collides with ball $X$ again. After this collision the speed of ball $Y$ is twice the speed of ball $X$ and the speed of ball $Z$ is four times the speed of ball $Y$, with the balls all travelling in the same direction.

Balls $X,Y$ and $Z$ lie at rest on a smooth horizontal surface, with $Y$ between $X$ and $Z$. Balls $X$ and $Z$ each have mass $2 \mathrm{kg}$ and ball $Y$ has mass $1 \mathrm{kg}$. Ball $X$ is given a velocity of $1 \mathrm{m}{ \mathrm{s}}^{-1}$ towards ball $Y$, Balls $X$ and $Y$ collide. After this collision the speed of ball $X$ is $0.4{ \mathrm{ms}}^{-1}$ in its original direction.

Work out the loss in kinetic energy in this impact.