A 0.0010-kg pellet is fired at a speed of 50.0m/s at a motionless 0.35-kg piece of balsa wood. When the 
pellet hits the wood, it sticks in the wood and they slide off together. With what speed do they slide?

Answer: The pellet and wood slide at a speed of 0.142 m/s.Explanation: To calculate the velocity of the pellet and wood after the collision, we use the equation of law of conservation of momentum, which is:[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]where,[tex]m_1[/tex] = mass of pellet = 0.001kg[tex]u_1[/tex] = Initial velocity of pellet = 50m/s[tex]v_1[/tex] = Final velocity of pellet = v m/s[tex]m_2[/tex] = mass of wood = 0.36kg[tex]u_2[/tex] = Initial velocity of wood = 0m/s[tex]v_2[/tex] = Final velocity of wood = v m/sPutting values in above equation, we get:[tex](0.001\times 50)+(0.35\times 0)=(0.001+0.35)v\\\\v=0.142m/s[/tex]Hence, the pellet and wood slide at a speed of 0.142 m/s.

p = m*vconservation of momentum suggestsinitial momentum equals final momentummv-initial = mv-final(0.0010 kg)(50 m/s) = (0.0010 kg + 0.35 kg)vthus:v = (0.0010)(50)/(0.351) = 0.142 m/s