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Difficulty: MediumNewton's Laws of Motion and Linear Momentum

A trolley of mass 0.40 kg0.40\text{ kg} moving to the right with a velocity of 5.0 m s−15.0\text{ m s}^{-1} collides head-on with a second trolley of mass 0.60 kg0.60\text{ kg} moving to the left at 2.5 m s−12.5\text{ m s}^{-1}. If the two trolleys stick together on impact, what is their combined velocity immediately after the collision?

  1. 0.50 m s−10.50\text{ m s}^{-1} to the rightAnswer
  2. B
    3.50 m s−13.50\text{ m s}^{-1} to the right
  3. C
    2.50 m s−12.50\text{ m s}^{-1} to the right
  4. D
    1.25 m s−11.25\text{ m s}^{-1} to the right

Answer

0.50 m s−10.50\text{ m s}^{-1} to the right
By adopting a standard vector sign convention where movement to the right is positive and movement to the left is negative, the initial momentum is pi=(0.40 kg×5.0 m s−1)+(0.60 kg×−2.5 m s−1)=2.0−1.5=0.50 kg m s−1p_i = (0.40 \text{ kg} \times 5.0 \text{ m s}^{-1}) + (0.60 \text{ kg} \times -2.5 \text{ m s}^{-1}) = 2.0 - 1.5 = 0.50 \text{ kg m s}^{-1}. Because the trolleys stick together, their total mass becomes 0.40 kg+0.60 kg=1.0 kg0.40 \text{ kg} + 0.60 \text{ kg} = 1.0 \text{ kg}. Dividing the total momentum by the combined mass gives a final velocity of +0.50 m s−1+0.50 \text{ m s}^{-1}, indicating movement to the right.

Step-by-Step Solution

1
Assign a directional sign convention for velocity vectors
Rightward velocity u1=+5.0 m s−1u_1 = +5.0\text{ m s}^{-1}, Leftward velocity u2=−2.5 m s−1u_2 = -2.5\text{ m s}^{-1}
Linear momentum is a vector quantity, so direction must be accounted for using opposite algebraic signs.
2
Calculate total initial linear momentum of the system
pi=m1u1+m2u2=(0.40×5.0)+(0.60×(−2.5))=2.0−1.5=+0.50 kg m s−1p_i = m_1 u_1 + m_2 u_2 = (0.40 \times 5.0) + (0.60 \times (-2.5)) = 2.0 - 1.5 = +0.50\text{ kg m s}^{-1}
Sum the individual initial momenta of both trolleys.
3
Apply the principle of conservation of linear momentum to solve for final combined velocity
v=pim1+m2=+0.500.40+0.60=+0.50 m s−1v = \frac{p_i}{m_1 + m_2} = \frac{+0.50}{0.40 + 0.60} = +0.50\text{ m s}^{-1}
Since no external net force acts on the system, total initial momentum equals total final momentum.

Key Concept

Conservation of Linear Momentum in Inelastic Collisions
Estimated Time:1m 30s
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Topics

Newton's Laws of Motion and Linear Momentum

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