Find v2v1v2v1, the ratio of the speed v2v2v_2 of the two-block system after the first collision to the speed v1v1v_1 of the block of mass m1m1m_1 before the collision.

v₂/v₁ = 0.777

Note: The question is incomplete. The complete question is given below;

A block of mass m1 1.40 kg moving at 12.00 m/s undergoes a completely inelastic collision with a stationary block of mass m2 0.400 kg . The blocks then move, stuck together, at speed v2. After a short time, the two-block system collides inelastically with a third block, of mass m3 2.70 kg, which is initially at rest. The three blocks then move, stuck together, with speed v3.(Figure 1) Assume that the blocks slide without friction Express your answer numerically using three significant figures

Part A Find , the ratio of the velocity 2 of the two-block system after the first collision to the velocity V1 of the block of mass m1 before the collision.

Explanation:

Velocity, v₁ (of mass m₁) = 12.0 m/s

Velocity of the two masses after inelastic collision = v₂

According to the principle of conservation of linear momentum, momentum before collision is equal to momentum after collision

m₁u₁ + m₂u₂ = (m₁ + m₂)v₂

where m₁ = mass of first block = 1.40 kg

u₁ = initial velocity of first block = 12.0 m/s

m₂ = mass of second block = 0.40 kg

u₂ = initial velocity of second block = 0 m/s

v₂ = final velocity of the two-block system = ?

Substituting the values in the conservation of momentum formula

1.4 * 12 + 0.4 * 0 = (1.4 + 0.4) * v₂

16.8 + 0 = 1.8 * v₂

v₂ = 16.8/1.8

v₂ = 9.33 m/s

v₂/v₁ = 9.33/12.0

v₂/v₁ = 0.777

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