Question

On a frictionless track, Cart A (velocity = +v; momentum = +10 N-s) collides with Cart B which is initially motionless. Cart A bounces off and heads back in the same direction from which it came, with velocity now equal to − ଵ ଷ v. (Hint: Cart A’s final momentum is negative.) Write down the specified numerical values of momentum in units of N-s. (Include correct +/− signs!)

A: initial momentum: _______ N-s; final momentum: __________ N-s; change in momentum: _________ N-s

B: initial momentum: _______ N-s; final momentum: __________ N-s; change in momentum: _________ N-s

Answer 1

Since in this collision the body A turns back on its path with exactly the same velocity as before the collision, this is a Case of elastic collision I.e both kinetic energy and momentum will be conserved for the whole system.

Now before the initial momentum of A was +10N-s and since it turns back on its path with exactly the same velocity, it's final momentum will be : - 10 N-s

Change in momentum will be= final momentum - initial momentum

= - 10 - (+10) = - 20 N-s

FOR the body B:

The initial momentum is 0 N-s

The final momentum will also be ZERO because due to momentum and energy conservation all the momentum is already given to the body A ( since it bounces back with the same velocity) therefore body B will remain at rest.

Change in momentum is also ZERO.