Answer to Question #350252 in Physics for mlas2

Question

A toy car having mass m = 1.10 kg

 collides inelastically with a toy train of mass M = 3.80 kg.

 Before the collision, the toy train is moving in the positive x-direction with a velocity of V_i = 2.45 m/s

 and the toy car is also moving in the positive x-direction with a velocity of v_i = 4.95 m/s.

 Immediately after the collision, the toy car is observed moving in the positive x-direction with a velocity of 2.25 m/s.

(a) Determine V_f, the final velocity of the toy train.

 

(b) Determine the change ΔKE

 in the total kinetic energy. Assume friction and the rotation of the wheels are not important so that they do not affect ΔKE.

Answer 1

(a) The law of conservation of momentum says

$$mv_i+MV_i=mv_f+MV_f$$

So

$$V_f=\frac{m}{M}(v_i-v_f)+V_i\\ =\frac{1.10}{3.80}(4.95-2.25)+2.45=3.23\:\rm m/s$$

(b)

$$\Delta KE=(\frac{mv_i^2}{2}+\frac{MV_i^2}{2})-(\frac{mv_f^2}{2}+\frac{MV_f^2}{2})$$

$$=(\frac{1.10*4.95^2}{2}+\frac{3.80*2.45^2}{2})-(\frac{1.10*2.25^2}{2}+\frac{3.80*3.23^2}{2})$$

$$=2.27\:\rm J$$