Question

(i) What is the velocity of the object just before and just after it collides with the wall. (ii) How far is the original starting point of the object from the wall, and how far from the wall does it travel before stopping after colliding with the wall?

A block slides toward a wall. The frictional force between the block and the floor is `0.12` `N`. Its initial velocity is `3.0` `ms^-1`, & it strikes the wall after `2.0` `s`. It collides with the wall & loses `0.072` `J` of energy.

Answer 1

(i) `v=1.4` `ms^-1` just before the mass hits the wall, and `1.0` `ms^-1` just after. The time taken until the block comes to rest is `2.8` s. (You draw the graph.)
(ii) From the start point to the wall is `4.4` `m` and from the wall to the stopping point is `0.63` `m`.

Explanation:

The acceleration of the block due to the frictional force will be given by:

`a=F/m=0.12/0.15=0.8` `ms^-2`

Since this acceleration is in a direction opposite to the initial velocity, it will have a negative sign.

Its initial velocity, `u=3.0` `ms^-1`, and its velocity after `2.0` `s`, just before it strikes the wall, will be:

`v=u+at=3+(-0.8)xx2=1.4` `ms^-1`

From this we can calculate that its kinetic energy just before the collision will be:

`E_k=1/2mv^2=1/2xx0.15x1.4^2=0.147` `J`

We are told that it loses `0.072` `J` of energy in the collision, so its kinetic energy immediately after the collision will be `0.147-0.072=0.075` `J`.

From this we can calculate its velocity immediately after the collision:

`E_k=1/2mv^2`

`v=sqrt((2E_k)/m)=sqrt((2xx0.075)/0.15)=1` `ms^-1`

Since this velocity (after bouncing off the wall) is in the opposite direction to the initial velocity it will have a negative sign. At the same time, though, because the acceleration is due to friction and will always act in the opposite direction to the motion, it will now have a positive sign.

We want to know the time taken for the object to come to rest. When that occurs, the final velocity will be zero:

`v=u+at`

`t=(v-u)/a=(0-0.8)/-1=0.8` `s`

Add this to the `2.0` `s` taken for the mass to reach the wall and the mass comes to a halt at `t=2.8` `s`.

Note that the velocity starts positive, decreases and becomes negative before going to zero.

We have now answered all of part `(i)` of the question.

To find the distance from the start point to the wall, we use:

`v^2=u^2+2as`

Rearranging, `s=(v^2-u^2)/2a=(1.4^2-3^2)/(2xx(-0.8))=4.4` `m`

We can do the same calculation for the distance from the wall to `Z`, the place where the block comes to rest.

`s=(v^2-u^2)/2a=(0^2-1^2)/(2xx(0.8))=-0.63` `m`

That is, `0.63` `m` in the direction opposite to the original direction of motion.