Question

Difficulty: HardNewton's Laws of Motion and Linear Momentum

A shell of total mass 5.0 kg5.0\text{ kg} is moving horizontally with a velocity of 20 m s120\text{ m s}^{-1} when an internal explosion splits it into two fragments. One fragment of mass 2.0 kg2.0\text{ kg} is propelled backward along the original path at a speed of 10 m s110\text{ m s}^{-1}. What is the magnitude of the velocity, in m s1\text{m s}^{-1}, of the second fragment immediately after the explosion?

Answer: 40 m s^-1

Answer

The magnitude of the velocity of the second fragment immediately after the explosion is 40 m s140\text{ m s}^{-1}.
According to the Law of Conservation of Linear Momentum, the total momentum of a system remains constant when no net external horizontal force acts on it. Taking the original direction of motion as positive, the initial momentum is 100 kg m s1100\text{ kg m s}^{-1}. Since the 2.0 kg2.0\text{ kg} fragment moves backward at 10 m s110\text{ m s}^{-1}, its momentum is 20 kg m s1-20\text{ kg m s}^{-1}. For total momentum to remain +100 kg m s1+100\text{ kg m s}^{-1}, the remaining 3.0 kg3.0\text{ kg} fragment must carry a momentum of +120 kg m s1+120\text{ kg m s}^{-1}, which corresponds to a velocity of 40 m s140\text{ m s}^{-1}.

Step-by-Step Solution

1
Calculate the initial momentum of the shell prior to the explosion.
pi=5.0 kg×20 m s1=100 kg m s1p_i = 5.0\text{ kg} \times 20\text{ m s}^{-1} = 100\text{ kg m s}^{-1} in the initial forward direction.
Before the internal explosion, the system consists of a single mass moving with a constant velocity.
2
Apply the Law of Conservation of Linear Momentum taking vector direction into account.
pi=m1v1+m2v2    100=2.0(10)+3.0v2p_i = m_1 v_1 + m_2 v_2 \implies 100 = 2.0(-10) + 3.0 v_2
In the absence of external forces, total momentum is conserved. The fragment propelled backward takes a negative sign relative to the initial forward motion.
3
Solve the algebraic equation for the unknown velocity v2v_2.
100+20=3.0v2    120=3.0v2    v2=40 m s1100 + 20 = 3.0 v_2 \implies 120 = 3.0 v_2 \implies v_2 = 40\text{ m s}^{-1}
Isolating v2v_2 gives the forward velocity magnitude of the remaining 3.0 kg3.0\text{ kg} piece.

Key Concept

Conservation of Linear Momentum in Explosions (1D Vector Sign Convention)
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