A railroad handcar is moving along straight, frictionless tracks with negligible

Question

A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of 160 and is traveling east with a velocity of magnitude 5.30 . Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks. Determine the result in each case: 1) An object with a mass of 24.0 is thrown sideways out of the car with a speed of 2.50 relative to the car's initial velocity 2) An object with a mass of 24.0 is thrown backward out of the car with a velocity of 5.30 relative to the initial motion of the car. 3) An object with a mass of 24.0 is thrown into the car with a velocity of 6.10 relative to the ground and opposite in direction to the initial velocity of the car.

Explanation / Answer

Case 1: There is no reactive force in the east-west direction if the 26.0kgs were thrown out at an exact right angle. The velocity should not have changed. Case 2: Initial momentum is: p = mv p = 200kg x 5.50m/s[east] Initial momentum should equal final momentum which with two terms (two objects, 1 is the car, 2 is the other mass; let east be the positive direction) would be: p(final) = m1v1 + m2v2 (200kg)(5.50m/s) = (274kg)v1 + (26.0kg)(-5.50m/s) Solve for v1 and that's the car's velocity. Case 3: Same thing, but arranged a little differently. Find the initial momentum of the system. Two terms for either object. For ease of calculation, set one direction (east or west) as positive. p(initial) = m1v1 + m2v2 Conversation of Momentum dictates that p(init)=p(final). Since the thing was thrown into the car, they are now one object, one body--so there's only one mv term to worry about with the masses ADDED TOGETHER. p(final) = m(final, 1 and two)v(final) Solve for v(final).

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