Answer to Question #261890 in Physics for Charlie

Question

A 25 kg object with 3.0 m/s velocity travelling to the right hits object B, which has a mass of 15 kg and 6.0 m/s velocity moving and travelling to the left. After collision, determine the velocity of object A if object B:

a. Remains travelling to the left for 0.30 m/s

b. Recoils to the right at 0.45 m/s, and

c. Remains together with object A.

Answer 1

(a) Let's apply the law of conservation of momentum:

$$m_Av_{Ai}-m_Bv_{Bi}=m_Av_{Af}-m_Bv_{Bf},$$ $$v_{Af}=\dfrac{m_Av_{Ai}-m_Bv_{Bi}+m_Bv_{Bf}}{m_A},$$ $$v_{Af}=\dfrac{25\ kg\times3.0\ \dfrac{m}{s}-15\ kg\times6.0\ \dfrac{m}{s}+15\ kg\times0.30\ \dfrac{m}{s}}{25\ kg},$$ $$v_{Af}=-0.42\ \dfrac{m}{s}.$$

The sign minus means that the object A moves in the opposite direction after the collision (to the left).

(b) Let's apply the law of conservation of momentum:

$$m_Av_{Ai}-m_Bv_{Bi}=m_Av_{Af}+m_Bv_{Bf},$$ $$v_{Af}=\dfrac{m_Av_{Ai}-m_Bv_{Bi}-m_Bv_{Bf}}{m_A},$$ $$v_{Af}=\dfrac{25\ kg\times3.0\ \dfrac{m}{s}-15\ kg\times6.0\ \dfrac{m}{s}-15\ kg\times0.45\ \dfrac{m}{s}}{25\ kg},$$ $$v_{Af}=-0.87\ \dfrac{m}{s}.$$

The sign minus means that the object A moves in the opposite direction after the collision (to the left).

(c) Let's apply the law of conservation of momentum:

$$m_Av_{Ai}-m_Bv_{Bi}=(m_A+m_B)v_{common},$$ $$v_{common}=\dfrac{m_Av_{Ai}-m_Bv_{Bi}}{m_A+m_B},$$ $$v_{common}=\dfrac{25\ kg\times3.0\ \dfrac{m}{s}-15\ kg\times6.0\ \dfrac{m}{s}}{25\ kg+15\ kg},$$ $$v_{common}=-0.375\ \dfrac{m}{s}.$$

The sign minus means that the combination of objects A and B moves in the opposite direction after the collision (to the left).