Block 1, of mass m1 = 9.30kg . moves along a frictionless air track with speed v
ID: 2180546 • Letter: B
Question
Block 1, of mass m1 = 9.30kg . moves along a frictionless air track with speed v1 = 13.0m/s - It collides with block 2, of mass m2 = 59.0kg , which was initially at rest. The blocks stick together after the collision. Find the magnitude p, of the total initial momentum of the two-block system. Express your answer numerically. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically. What is the change delta K = Kfinal - Kinitial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules.Explanation / Answer
By the law of momentum m1v1 + m2v2 = m1u1 + m2u2 since the block 2 is at rest therefore after collision its momentum = 0 block 1 will be now moving with v2 and block 2 will be moving with v 9.3*13 + 0 = 9.3v +59vf total momentum of the system will be same 120 = 9.3vf +59v from here we can find the value of v in term of vf v = 120.9-59vf If v = 0 vf = 120/59 vf = 2.033 m/s