Block 1, of mass m_1 = 7.30 kg, moves along a frictionless air track with speed
ID: 1496894 • Letter: B
Question
Block 1, of mass m_1 = 7.30 kg, moves along a frictionless air track with speed upsilon_1 = 15.0 m/s. It collides with block 2, of mass m_2 = 59.0 kg, Which was intally at rest. The blocks stick together after the collision. Find the magnitude p_i of the total initial momentum of the two-block system. Express your answer numerically. Find upsilon_f, the magnitude of the final velocity of the two-block system. Express your answer numerically. What is the change DeltaK = K_final - K_initial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules.Explanation / Answer
m = 7.3 kg
M = 59 kg
Initial velocity of mass m is u = 15 m/s
Initial velocity of block M is U = 0
Velocities after collision v = ?
Initial momentum of the system P = mu +MU
= (7.3 x 15) +(59x0)
= 109.5 kgm/s
(b)From law of conservation of momentum ,
mu + MU = (m+M) v
109.5 = (7.3 +59) v
From this v = 1.651 m/s
(C). Initial kinetic energy of the system K = (1/2)mu 2+(1/2)MU 2
K = (1/2)(7.3)(15 2)
= 821.25 J
Final kinetic energy K ' = (1/2) (m+M) v 2
=(1/2)(7.3+59)(1.651) 2
= 90.42 J
Change in kinetic energy = K - K '
= 730.83 J