Block 1, of mass m1 = 6.30kg , moves along a frictionless air track with speed v
ID: 1282682 • Letter: B
Question
Block 1, of mass m1 = 6.30kg , moves along a frictionless air track with speed v1 = 31.0m/s . It collides with block 2, of mass m2 = 39.0kg , which was initially at rest. The blocks stick together after the collision.
Find the magnitude pi of the total initial momentum of the two-block system.
Find vf, the magnitude of the final velocity of the two-block system.
What is the change ?K=Kfinal?Kinitial in the two-block system's kinetic energy due to the collision?
Block 1, of mass m1 = 6.30kg , moves along a frictionless air track with speed v1 = 31.0m/s . It collides with block 2, of mass m2 = 39.0kg , which was initially at rest. The blocks stick together after the collision. Find the magnitude pi of the total initial momentum of the two-block system. Find vf, the magnitude of the final velocity of the two-block system. What is the change ?K=Kfinal?Kinitial in the two-block system's kinetic energy due to the collision?Explanation / Answer
a) pi = 6.3*31 = 195.3 kgm/s
b) vf = pi / (mi + mii ) = 195.3/ 45.3 = 4.31 m/s
c) ?KE = (0.5 * 45.3 * 4.31^2)