Question 1
Block A of mass m₁ = 8.0 kg is travelling initially at u₁ = 6.0 m/s in the positive x-direction. It collides with block B of mass m₂ = 12.0 kg, which is moving in the same direction with velocity u₂ = 3.0 m/s. If immediately after the collision, the velocity of block A is v₁ = 4.0 m/s in the positive x-direction, find the velocity of block B (v₂) immediately after the collision.
- A.4.33 m/s to the negative x-axis
- B.3.71 m/s to the negative x-axis
- C.3.71 m/s to the positive x-axis
- D.4.33 m/s to the positive x-axis
Show the answer and explanation
Answer: D 4.33 m/s to the positive x-axis
Why
No outside force acts on the two blocks, so the total momentum before the collision equals the total momentum after it: \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \).
Momentum before: \( (8.0)(6.0) + (12.0)(3.0) = 48 + 36 = 84 \, \text{kg·m/s} \).
Momentum after: \( (8.0)(4.0) + 12.0\,v_2 = 32 + 12v_2 \).
Set them equal: \( 84 = 32 + 12v_2 \), so \( 12v_2 = 52 \) and \( v_2 = 4.33 \, \text{m/s} \).
The result is positive, so block B keeps moving along the positive x-direction. Options A and B fail because a positive answer cannot point along the negative x-axis.
Textbook: Physics Grade 11, Unit 4 Dynamics, section 4.6 Impulse and Linear Momentum (page 148).