A bullet of mass m_b = 3.5 times 10^-3 kg is fired horizontally at two blocks M_
ID: 1419681 • Letter: A
Question
A bullet of mass m_b = 3.5 times 10^-3 kg is fired horizontally at two blocks M_1 = 1.2 kg and M_2 = 1.8 kg. the two blocks are sitting on a horizontal smooth surface. the bullet passes through M_1 and embeds itself on M_2. This results in M_1 moving to the right with a speed of 0.63 m/s and M_2 plus the bullet moving to the right with a speed of 1.4 m/s as shown in the diagram. What is the initial velocity of the bullet before encountering M_1 and M_2? What is the velocity of the bullet immediately after it emerges from the first block(M_1)?Explanation / Answer
a) Apply conservation of momentum
initial momentum of bullet = final momentum M1,M2 and bullet
mb*u1 = M1*v1 + (M2 + mb)*v2
u1 = (M1*v1 + (M2+mb)*v2)/mb
= (1.2*0.63 + (1.8+0.0035)*1.4)/0.0035
= 937.4 m/s <<<<<<----------Answer
b) when the bullet just emerges from M1
Apply conservation of momentum
initial momentum of bullet = final momentum M1 and bullet
mb*u1 = M1*v1 + mb*u1f
==> u1f = (mb*u1 - M1*v1)/mb
= (0.0035*937.4 - 1.2*0.63)/0.0035
= 721.4 m/s <<<<<<----------Answer
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