Question

In the figure, a bullet with mass m1 moving directly upward with speed v1i strikes and passes through the center of mass of a block with mass m2 which is initially at rest. The bullet emerges from the block moving directly upward and has slowed to a speed v1f. To what maximum height does the block then rise above its initial position? State your answer in terms of the given variables (use g where applicable)

Answer 1

Solution:

According to the conservation of momentum.
(m₁)(v_1i) = (m₁)(v_1f) + (m₂)(v₂)
v₂ = (m₁)(v_1i - v_1f) / (m₂)

Now : to find how high the block can go with initial velocity v₂:
Using equation : v² - (v₂)² = 2ay

Since, initial velocity of the block is Zero.

So : (v₂)² = 2gy
y = (v₂)² / (2g)

Putting values of v₂ from above equations, we get :
y = (m₁)²(v_1i - v_1f)² / [(2g)(m₂)²]