OCW Scholar - Physics I: Classical Mechanics
OpenStudy (anonymous):

Hello, how do I solve this problem? im confused, its about the concept cuestions on momentum and impulse. The first one ask about how is the impulse of the lighter car respect of the heavier car.But the problem is that if I calculate the momentum of the lighter car and the momentum of the heavier car, those are diferent because of the mass. And if I take the impulse as the integral of F(t)dt then the impulse is the same for both cars. Can anyone help me, please?

OpenStudy (anonymous):

The conceptual answers says that I must take the integral, but it makes me unconfortable because impulse=delta momentum but in this example they give me diferent results.

OpenStudy (anonymous):
OpenStudy (anonymous):

I'm not sure I understand what you are confused about, but I will attempt an answer all the same... Impulse is defined as the integral over F with respect to time, since F is constant this amounts to I = Ft which is the same for both masses. Now, since F = ma = m dv/dt = dp/dt = "change of momentum". It follows that the momentum of the two masses change by an equal amount, hence their momentum is the same at all instances of time. In fact, since the momentum is equal then we must have m1 v1 = m2 v2, hence the speed of the heavier mass m2 is half the speed of the lighter mass. This result can also be seen by just utilizing newtons's second law: Since F = ma we must have m1 a1 = m2 a2, hence a1/a2 = m2 /m1 = 2. Thus the acceleration of the lighter mass is two times the acceleration of the heavier mass. It follows then that the end speed of the lighter mass is two times the speed of the heavier mass (since both are accelerated the same amount of time). If we simply calculate the momentum at the end we will thus have: p1 = m1 v1 = m1 (2 v2) = 2 m1 v2 p2 = m2 v2 = (2 m1) v2 = 2 m1 v2 hence p1 = p2.