A block with a mass of m1 = 5 kg is released from rest and slides down a frictio

Question

A block with a mass of m1 = 5 kg is released from rest and slides down a frictionless incline and strikes a second mass, also initially at rest, with m2 = 3 kg. After the collision, the two masses stick and move off together with a speed of v = 1.5 m/s. The two masses then encounter a second frictionless incline and slide up that incline before eventually coming to rest.
a) At what speed was m1 travelling immediately before striking m2?
b) How high up the first incline did m1 start?
c) How high up the second incline do the two masses go before coming to rest?

Explanation / Answer

The model used to solve this problem will disregard friction and assume constant acceleration.

Part a can be solved using momentum: p=mv. Since momentum is conserved, p initial will be equal to p final. Therefore, the mass of the first block multiplied by its initial velocity will be equal to the total mass of the two blocks multiplied by their velocity of 1.5m/s:

m1v1 = (m1 + m2)v2

(5 kg)v1 = (5 kg + 3 kg)(1.5 m/s)

Solve for v1.

Part b can be solved using kinematics. The variable a will be acceleration, which in this case is gravitational acceleration (-9.81 m/s2). The variable y1 will represent the initial vertical height of the first block. Plug in the value found in part a for vi.

vf2 = vi2 + 2ay1

(1.5 m/s)2 = vi2 + 2(-9.81 m/s2)y1

Solve for y1.

Part c can also be solved using kinematics. The equation used to solve part b can also be used to determine how high the blocks will go before gravity pulls them back down. In this case, a will represent (gravitational) acceleration, y2 will represent the final height being found, the final velocity will be the blocks at rest (0 m/s), and the initial velocity will be the velocity at which the blocks are moving after they collide.

vf2 = vi2 + 2ay2

(0 m/s)2 = (1.5 m/s)2 + 2(-9.81 m/s2)y2

Solve for y2.

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