# Mechanics Problem; Any help would be appreciated

• Mar 26th 2011, 05:12 AM
osmosis786
Mechanics Problem; Any help would be appreciated
The details of the question are below;

Quote:

Identical Cars A, B and C, are initially at rest in a straight line on a horizontal surface with their brakes off. Car A is pushed towards the others so that it hits Car B with speed ${2ms^-1}$.
Given that the coefficient of restitution between any two cars is 0.75, find the speeds of the cars after all collisions have finished.
Spoiler:
A=0.223, B=0.246, C=1.53

Please bear in mind that this is college work (16-18) and so try to keep the explanations limited to that as much as possible.

• Mar 26th 2011, 01:40 PM
Opalg
Quote:

Originally Posted by osmosis786
Identical Cars A, B and C, are initially at rest in a straight line on a horizontal surface with their brakes off. Car A is pushed towards the others so that it hits Car B with speed $2ms^{-1}$.
Given that the coefficient of restitution between any two cars is 0.75, find the speeds of the cars after all collisions have finished.

Suppose that X and Y are two objects with masses $M_X$ and $M_Y$, moving along a line with velocities $u_X$ and $u_Y$. When one of them shunts into the other, they separate, moving at new speeds $v_X$ and $v_Y$. You need two equations to work out the new speeds after the collision.

The first equation is given by conservation of momentum: $M_Xv_X + M_Yv_Y = M_Xu_X + M_Yv_Y$. If both objects have the same mass (as in the case of these cars) then the M's all cancel, and you are left with the conservation of momentum equation $v_X+v_Y = u_X+u_Y.$

The second equation is the restitution equation, which says that the coefficient of restitution is equal to $\dfrac{v_Y-v_X}{u_X-u_Y}$. In this problem, the coefficient is 0.75, so the equation becomes $v_Y-v_X = 0.75(u_X-u_Y).$

Now, what happens in these car pile-ups? The first collision occurs when car A, travelling at 2m/s, hits car B. If the velocities after this collision are $v_A$ and $v_B$, then the momentum equation says that $v_A+v_B = 2+0 = 2$. The restitution equation says that $v_B-v_A = 0.75(2-0) = 1.5.$ That gives you two simultaneous equations for $v_A$ and $v_B$. Solve them. You should find that $v_A = 0.25$ and $v_B = 1.75.$

The next thing that happens is that car B, travelling at 1.75m/s, hits the stationary car C. You can go through the same procedure as for the first collision, solving two simultaneous equations to find the velocities $v_B'$ and $v_C$ after the collision. I made them $v_B' = 0.219$ and $v_C = 1.531$.

But now car A, still moving at 0.25m/s, is going faster than car B, moving at 0.219m/s. So car A will catch up with car B and there will be another collision. Once again, go through the whole procedure of solving two simultaneous equations to find the velocities $v_A'$ and $v_B''$ after this collision. You should get $v_A' = 0.223$ and $v_B'' = 0.246.$ At this stage, each car is going slower than the one in front of it, so there will be no more collisions and the calculation is complete.
• Mar 27th 2011, 03:39 AM
osmosis786
Thanks a lot. I see where I was going wrong now.

Kudos to you.