Question

A 4.25-g bullet traveling horizontally with a velocity of magnitude 375 m/s is fired into a wooden block with mass 1.12 kg, initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speed reduced to 114 m/s. Part A How fast is the block moving just after the bullet emerges from it

Answer 1

0.99 m/s

Step-by-step explanation:

From the question,

Note: The collision between the bullet the the wooden block is elastic.

Total momentum before collision = Total momentum after collision.

mu+Mu' = mv+Mv'................ Equation 1

Where m = mass of the bullet, u = initial velocity of the bullet, M= mass of the wooden block, u' = initial velocity of the wooden block, v = final velocity of the bullet, v' = final velocity of the wooden block

Since, u' = 0 m/s

Therefore,

mu = mv+Mv'

make v' the subject of the equation

v' = (mu-mv)/M.................... Equation 2

Given: m = 4.25 g = 0.00425 kg, u = 375 m/s, M = 1.12 kg, v = 114 m/s.

Substitute into equation 2

v' = [(0.00425×375)-(0.00425×114)]/1.12

v' = (1.59375-0.4845)/1.12

v' = 0.99 m/s

Answer 2

0.9904 m/s

Step-by-step explanation:

To solve this problem we need to use the conservation of momentum:

m1v1 + m2v2 = m1'v1' + m2'v2'

Using this equation, we have:

m_bullet*v_bullet = m_bullet_after*v_bullet_after + m_block*v_block

0.00425 * 375 = 0.00425 * 114 + 1.12 * v_block

1.12 * v_block = 1.5938 - 0.4845

1.12 * v_block = 1.1093

v_block = 1.1093 / 1.12

v_block = 0.9904 m/s