# 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 \$\displaystyle {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 \$\displaystyle 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 \$\displaystyle M_X\$ and \$\displaystyle M_Y\$, moving along a line with velocities \$\displaystyle u_X\$ and \$\displaystyle u_Y\$. When one of them shunts into the other, they separate, moving at new speeds \$\displaystyle v_X\$ and \$\displaystyle v_Y\$. You need two equations to work out the new speeds after the collision.

The first equation is given by conservation of momentum: \$\displaystyle 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 \$\displaystyle 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 \$\displaystyle \dfrac{v_Y-v_X}{u_X-u_Y}\$. In this problem, the coefficient is 0.75, so the equation becomes \$\displaystyle 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 \$\displaystyle v_A\$ and \$\displaystyle v_B\$, then the momentum equation says that \$\displaystyle v_A+v_B = 2+0 = 2\$. The restitution equation says that \$\displaystyle v_B-v_A = 0.75(2-0) = 1.5.\$ That gives you two simultaneous equations for \$\displaystyle v_A\$ and \$\displaystyle v_B\$. Solve them. You should find that \$\displaystyle v_A = 0.25\$ and \$\displaystyle 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 \$\displaystyle v_B'\$ and \$\displaystyle v_C\$ after the collision. I made them \$\displaystyle v_B' = 0.219\$ and \$\displaystyle 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 \$\displaystyle v_A'\$ and \$\displaystyle v_B''\$ after this collision. You should get \$\displaystyle v_A' = 0.223\$ and \$\displaystyle 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.