Question

Block 1, of mass m1m1m_1 = 6.70 kgkg , moves along a frictionless air track with speed v1v1v_1 = 27.0 m/sm/s . It collides with block 2, of mass m2m2m_2 = 57.0 kgkg , which was initially at rest. The blocks stick together after the collision. (Figure 1)

Figure

1 of 1The figure shows two states of a system of two blocks, labeled 1 and 2, of masses m 1 and m 2, respectively. Block 2 is to the right of block 1. In state labeled “before collision”, block 1 moves to the right with velocity v 1 and block 2 is at rest. In state labeled “after collision”, the blocks move together with velocity v f directed to the right.

Part A: Find the magnitude pi of the total initial momentum of the two-block system.

B: Find vf, the magnitude of the final velocity of the two-block system.

C: What is the change ΔK=Kfinal−KinitialΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision?

Answer 1

given
m1 = 6.70 kg
v1 = 27.0 m/s
m2 = 57.0 kg
v2 = 0

A) the magnitude pi of the total initial momentum of the two-block system, Pi = m1*v1 + m2*v2

= 6.7*27 + 57*0

= 181 kg.m/s <<<<<<<<<---------------Answer

B) Apply conservation of momentum

momentum after the collsion = momentum before the collision
Pf = Pi

(m1 + m2)*vf = Pi

vf = Pi/(m1 + m2)

= 181/(6.7 + 57)

= 2.84 m/s <<<<<<<<<---------------Answer

c) delta_K = K_final - K_initial

= (1/2)*(m1 + m2)*vf^2 - (1/2)*m1*v1^2

= (1/2)*(6.7 + 57)*2.84^2 - (1/2)*6.7*27^2

= -2185 J <<<<<<<<<---------------Answer