Question

# Block 1, of mass m1 = 1.70 kg , moves along a frictionless air track with...

Block 1, of mass m1 = 1.70 kg , moves along a frictionless air track with speed v1 = 29.0 m/s . It collides with block 2, of mass m2 = 59.0 kg , which was initially at rest. The blocks stick together after the collision. (Figure 1)

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 ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision?

Mass of block 1 = m1 = 1.7 kg

Mass of block 2 = m2 = 59 kg

Velocity of block 1 before the collision = V1 = 29 m/s

Velocity of block 2 before the collision = V2 = 0 m/s

Velocity of the blocks after the collision = Vf

Initial momentum of the two block system = Pi

Pi = m1V1 + m2V2

Pi = (1.7)(29) + (59)(0)

Pi = 49.3 kg.m/s

By conservation of linear momentum,

m1V1 + m2V2 = (m1 + m2)Vf

(1.7)(29) + (59)(0) = (1.7 + 59)Vf

Vf = 0.812 m/s

Initial kinetic energy = KEi

KEi = m1V12/2 + m2V22/2

KEi = (1.7)(29)2/2 + (59)(0)2/2

KEi = 714.85 J

Final kinetic energy = KEf

KEf = (m1 + m2)Vf2/2

KEf = (1.7 + 59)(0.812)2/2

KEf = 20.01 J

Change in kinetic energy = KE

KE = KEf - KEi

KE = 20.01 - 714.85

KE = -694.84 J

Total initial momentum of the two block system = 49.3 kg.m/s

Final velocity of the two block system = 0.812 m/s

Change in kinetic energy due to the collision = - 694.84 J

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