Question

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 *p*i of the total initial momentum of
the two-block system.

Find *v*f, the magnitude of the final velocity of the
two-block system.

What is the change Δ*K*=*K*final−*K*initial
in the two-block system's kinetic energy due to the collision?

Answer #1

Mass of block 1 = m_{1} = 1.7 kg

Mass of block 2 = m_{2} = 59 kg

Velocity of block 1 before the collision = V_{1} = 29
m/s

Velocity of block 2 before the collision = V_{2} = 0
m/s

Velocity of the blocks after the collision = V_{f}

Initial momentum of the two block system = P_{i}

P_{i} = m_{1}V_{1} +
m_{2}V_{2}

P_{i} = (1.7)(29) + (59)(0)

P_{i} = 49.3 kg.m/s

By conservation of linear momentum,

m_{1}V_{1} + m_{2}V_{2} =
(m_{1} + m_{2})V_{f}

(1.7)(29) + (59)(0) = (1.7 + 59)V_{f}

V_{f} = 0.812 m/s

Initial kinetic energy = KE_{i}

KE_{i} = m_{1}V_{1}^{2}/2 +
m_{2}V_{2}^{2}/2

KE_{i} = (1.7)(29)^{2}/2 +
(59)(0)^{2}/2

KE_{i} = 714.85 J

Final kinetic energy = KE_{f}

KE_{f} = (m_{1} +
m_{2})V_{f}^{2}/2

KE_{f} = (1.7 + 59)(0.812)^{2}/2

KE_{f} = 20.01 J

Change in kinetic energy = KE

KE
= KE_{f} - KE_{i}

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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