Block 1, of mass m 1 = 4.70 kg , moves along a frictionless air track with speed
ID: 1547796 • Letter: B
Question
Block 1, of mass m1 = 4.70 kg , moves along a frictionless air track with speed v1 = 25.0 m/s . It collides with block 2, of mass m2 = 53.0 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.
Part B
Find vf, the magnitude of the final velocity of the two-block system.
Express your answer numerically.
m/s
Part C
What is the change K=KfinalKinitial in the two-block system's kinetic energy due to the collision?
Express your answer numerically in joules.
pi = kgm/s Before collision ms m1 After collisionExplanation / Answer
m1 = 4.70 kg
Initial speed of m1 is v1 = 25.0 m/s
mass m2 = 53.0 kg
Initial speed of m2 is v 2 = 0
Speed after collision V = ?
the magnitude pi of the total initial momentum of the two-block system = m1v1+m2v2
= 4.7(25) +53(0)
= 117.5 kgm/s
From law of conservation of momentum , m1v1+m2v2 = (m1+m2) V
4.7(25) +53(0) = (4.7+53) V
117.5 = 57.7 V
V = 2.036 m/s
the change K=KfinalKinitial in the two-block system's kinetic energy due to the collision is
= [(1/2)m1v1 2 +(1/2)m2v2 2 ] -[(1/2)(m1+m2)V 2]
= [(0.5x4.7x25 2 + 0] -[0.5x57.7x2.036 2 ]
= 1468.75 -119.6
= 1349.158 J