Block 1, of mass m1 = 7.90 kg , moves along a frictionless air track with speed

Question

Block 1, of mass m1 = 7.90 kg , moves along a frictionless air track with speed v1 = 25.0 m/s . It collides with block 2, of mass m2 = 23.0 kg , which was initially at rest. The blocks stick together after the collision. (Figure 1)

Part A Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically. pi = kg?m/s

Part Part B Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically. vf = m/s

Part C What is the change ?K=Kfinal?Kinitial in the two-block system's kinetic energy due to the collision?

Explanation / Answer

by using mument conversion

m1v1+m2v2=(m1+m2)Vf

7.90X25=30.9Vf

Vf=6.391m/s

initial energy of system is =kinetic energy of m1+kinetic energy of m2

initial energy=2468.75

initial energy = final energy

=final energy-initial energy

=631.0534-2468.75

=1837.69 loss

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