Question

A runaway 185 kg wheeled skip is rolling east along a road with a velocity of +2.2 m s−1. A 2560 kg car moving to the west with a velocity of −9.7 m s−1 hits the skip.
After the collision the skip is moving west with a velocity of −14.4 m s−1.
What is the velocity of the car after the collision? (in m s−1 to 2 s.f)

Answer 1

Given: $m_s=1.85\cdot 10^2 kg$ - mass of wheeled skip; $v_s=2.2m/s$ - velocity of skip before the collision; $V_s=-14.4 m/s$ - velocity of skip after the collision; $M_c=2.56\cdot 10^3 kg$ - mass of the car; $v_c=-9.7m/s$ - velocity of car before the collision. The velocity of the car after the collision $V_c$ we should to determine.

Solution: We should to apply the law of conservation of momentum. The momentum of the system before the collision is

(1) $P_0=m_s\cdot v_s+M_c\cdot v_c$

The momentum of the system after the collision is

(2) $P_1=m_s\cdot V_s+M_c\cdot V_c$

According to the law of conservation of momentum $P_0=P_1$ that is

(3) $m_s\cdot v_s+M_c\cdot v_c=m_s\cdot V_s+M_c\cdot V_c$

From (3) we have for the velocity of the car after the collision

(4) $V_c=(v_s-V_s)\frac{m_s}{M_c}+v_c=(2.2+14.4)(m/s)\frac{1.85\cdot 10^2kg}{2.56\cdot 10^3kg}-9.7m/s=\\=16.6\cdot 0.0723-9.7\simeq1.20-9.7=-8.50m/s$

The sign plus we used for the moving to the east, and the negative sign for the moving to the west. Thus the car after the collision continues move to the west with slightly smaller speed.

Answer: The velocity of the car after the collision is $-8.50 m/s$ to the west.