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Difficulty: HardNewton's Laws of Motion and Linear Momentum

A body AA of mass 4.0 kg4.0\text{ kg} moving due east at a velocity of 6.0 m s−16.0\text{ m s}^{-1} collides head-on with a body BB of mass 2.0 kg2.0\text{ kg} moving due west at 3.0 m s−13.0\text{ m s}^{-1}. If body AA continues to move due east after the collision with a speed of 1.0 m s−11.0\text{ m s}^{-1}, what is the velocity of body BB after the collision?

  1. 7.0 m s−17.0\text{ m s}^{-1} due EastAnswer
  2. B
    13.0 m s−113.0\text{ m s}^{-1} due East
  3. C
    7.0 m s−17.0\text{ m s}^{-1} due West
  4. D
    11.0 m s−111.0\text{ m s}^{-1} due East

Answer

7.0 m s−17.0\text{ m s}^{-1} due East
According to the principle of conservation of linear momentum, the total momentum before collision equals the total momentum after collision. Assigning positive to East and negative to West, total initial momentum is 4(6)+2(−3)=18 kg m s−14(6) + 2(-3) = 18\text{ kg m s}^{-1}. Equating this to final momentum 4(1)+2vB4(1) + 2v_B gives 2vB=142v_B = 14, resulting in vB=+7.0 m s−1v_B = +7.0\text{ m s}^{-1}, which represents 7.0 m s−17.0\text{ m s}^{-1} due East.

Step-by-Step Solution

1
Define a reference direction for 1D momentum vectors
Let East be positive (+) and West be negative (-).
Linear momentum is a vector quantity, so opposite directions must have opposite algebraic signs.
2
Calculate the total initial momentum before collision (pip_i)
pi=mAuA+mBuB=(4.0 kg)(+6.0 m s−1)+(2.0 kg)(−3.0 m s−1)=24.0−6.0=18.0 kg m s−1p_i = m_A u_A + m_B u_B = (4.0\text{ kg})(+6.0\text{ m s}^{-1}) + (2.0\text{ kg})(-3.0\text{ m s}^{-1}) = 24.0 - 6.0 = 18.0\text{ kg m s}^{-1}.
Body B moves west, so its initial velocity is −3.0 m s−1-3.0\text{ m s}^{-1}.
3
Formulate the total final momentum after collision (pfp_f)
pf=mAvA+mBvB=(4.0 kg)(+1.0 m s−1)+(2.0 kg)vB=4.0+2.0vBp_f = m_A v_A + m_B v_B = (4.0\text{ kg})(+1.0\text{ m s}^{-1}) + (2.0\text{ kg})v_B = 4.0 + 2.0 v_B.
Body A moves east after collision, so its final velocity is +1.0 m s−1+1.0\text{ m s}^{-1}.
4
Apply the Law of Conservation of Linear Momentum (pi=pfp_i = p_f)
18.0=4.0+2.0vB  ⟹  2.0vB=14.0  ⟹  vB=+7.0 m s−118.0 = 4.0 + 2.0 v_B \implies 2.0 v_B = 14.0 \implies v_B = +7.0\text{ m s}^{-1}.
Since total initial momentum equals total final momentum in an isolated system.
5
Interpret the sign of the calculated velocity
Since vBv_B is positive (+7.0 m s−1+7.0\text{ m s}^{-1}), body B moves at 7.0 m s−17.0\text{ m s}^{-1} due East.
Positive values correspond to the defined East direction.

Key Concept

Principle of Conservation of Linear Momentum in 1D head-on collisions
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Topics

Newton's Laws of Motion and Linear Momentum

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