Block 1, of mass m_1 = 6.70 kg , moves along a frictionless air track with speed
ID: 1348415 • Letter: B
Question
Block 1, of mass m_1 = 6.70 kg , moves along a frictionless air track with speed v_1 = 25.0 m/s . It collides with block 2, of mass m_2 = 39.0 kg , which was initially at rest. The blocks stick together after the collision. (Figure 1) Find the magnitude P_i of the total initial momentum of the two-block system. Express your answer numerically. Find v_f, the magnitude of the final velocity of the two-block system. Express your answer numerically. What is the change Delta K = K_final - K_initial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules.Explanation / Answer
A) initial momentum = m1v1 + mv2
Pi = (6.70 x 25) + ( 39 x 0) = 167.5 kg m/s
B) from momentum conservation,
initial momentum = final momentum
167.5 = (6.70 + 39)vf
vf =3.66 m/s ............Ans
C) Change in KE = final KE - initial KE
= [ 6.70 x 25^2 /2 ] - [ ( 6.70 + 39) x 3.66^2 / 2]
= 1787.66 J