A bullet with a mass 𝑚b=13.5 g is fired into a block of wood at velocity 𝑣b=245 m/s. The block is attached to a spring that has a spring constant 𝑘 of 205 N/m.205 The block and bullet continue to

Question

A bullet with a mass 𝑚b=13.5 g is fired into a block of wood at velocity 𝑣b=245 m/s. The block is attached to a spring that has a spring constant 𝑘 of 205 N/m.205 The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass 𝑚w of the wooden block.

Answer 1

Let

m_b = 13.5 g = 0.0135 kg be the mass of the bullet,

vb = 245 m/s be the velocity of the bullet,

m (to be computed) be the mass of the wooden block,

v (unknown) be the initial velocity of the block with the bullet lodged in,

k = 205 N/m be the spring constant,

s = 35 cm = 0.35 m be the compression of the spring.

There are two events going on:

- the bullet enters the wooden block,

- the block with the bullet compresses the spring.

When the bullet enters the block, energy is not conserved.

(We know that because conservation of energy would imply then the bullet bounces off the block.)

On the other hand, momentum is conserved.

In contrast, during the second event of compression of the spring, energy is conserved.

Initial momentum of the bullet before entering the wooden block:

m_bv_b

Momentum of the block with the bullet lodged in just before the spring starts compressing:

(m + m_b)v

Energy of the the whole system before the spring starts compressing:

(m + m_b)v²/2 (This is the kinetic energy of the block. The spring has energy 0.)

Energy of the system after the spring is compressed:

ks²/2 (This is the spring energy. The block has kinetic energy 0.)

So we have two equations:

m_bv_b = (m + m_b)v (conservation of momentum during the first event) (1)

(m + m_b)v²/2 = ks²/2 (conservation of energy during second event) (2)

From (1) express

v = m_bv_b/(m + m_b) (3)

Substitute (3) into (2)

(m + m_b)(m_bv_b/(m + m_b))²/2 = ks²/2 (4)

Simplify (4)

(m_bv_b)²/(m + m_b) = ks²

Express

m = (m_bv_b/s)²/k - m_b

Substitute actual numbers

m = (0.0135×245/0.35)²/205 - 0.0135 = 0.42 kg