A 0.11 kg steel bullet is shot at 210 m/s into a 2.3 kg stationary block of lead. After the collision, the bullet remains lodged inside the lead.

Question

The specific heats of steel and lead are 450 J/(kg⋅∘C) and 130 J/(kg⋅∘C)

a) How fast, in m/s, is the block of lead moving after the collision?

b) If the bullet is initially at 68∘C and the block is initially at 20∘C, what is the temperature of the block, in ∘C, just after the collision? Assume the bullet and the block reach thermal equilibrium essentially immediately during the collision.

Answer 1

The heat added to the block will come from two forms:

1) The heat added from the fact the bullet is 68° C and the block is 20°C

2) The heat added from the conversion of kinetic energy into heat energy.

Calculating the first:

If a bullet with a temperature of 68°C were just placed into a hole in the block we could calculate the heat added:

The heat from the bullet lost from the bullet would equal the heat added to the block except negative:

Q_bullet = -Q_block

Using Q = mCΔT where "m" is the mass of the object, "C" is the specific heat of the object, and ΔT is the final temp minus the initial temp:

Q_bullet = m_bulletC_bullet(T_f-bullet - T_i-bullet)

Q_bullet = (0.11)(450)(T_f-bullet - 68)

Q_bullet = (49.5)(T_f-bullet - 68)

Q_bullet = 49.5T_f-bullet - 3366

And, for the block:

Q_block = m_blockC_block(T_f-block - T_i-block)

Q_block = (2.3)(130)(T_f-block - 20)

Q_block = (299)(T_f-block - 20)

Q_block = 299T_f-block - 5980

Since Q_bullet = -Q_block then 49.5T_f-bullet - 3366 = -(299T_f-block - 5980)

We can also say that T_f-bullet = T_f-block since everything will be at the final temp so:

49.5T_f - 3366 = -(299T_f - 5980)

49.5T_f - 3366 = -299T_f + 5980

348.5T_f = 9346

T_f = 26.8 °C

Plugging this into Q_block = (2.3)(130)(T_f-block - 20), we get:

Q_block = (2.3)(130)(26.8 - 20) = 2039 joules

So 2039 joules is added to the block because of the hot temperature of the bullet.

Calculating the second:

Momentum is conserved so the momentum before the collision (P₁) is equal to the momentum after the collision (P₂)

P₁ = m_bullet•v_bullet = (0.11)(210) = 23,1 kg m/s

P₂ = (m_bullet + m_block)•v_f = (0.11 + 2.3)•v_f = 2.41v_f

Because P₁ = P₂ then 23.1 = 2.41v_f or v_f = 9.585 m/s

Energy is conserved and assuming that the kinetic energy before the collision is turned into a sum of kinetic energy and heat after the collision then:

E_K1 = E_K2 + Q_{added to block} or Q_{added to block} = E_K1 - E_K2

E_K1 = (1/2)(m_bullet)(v_bullet)² = (1/2)(0.11)(210)² = 2425.5 joules

E_K2 = (1/2)(m_bullet + m_block))(v_f)² = (1/2)(0.11 + 2.3))(9.585)² = 110.71 joules

So since Q_{added to block} = E_K1 - E_K2 then Q_{added to block} = 2425.5 - 110.71 = 2314.8 joules

Total energy added to the block = 2039 + 2314.8 = 4353 joules

Final temp of the block can be calculated from Q = mCΔT:

2314.8 = (2.3)(130)(T_f - 20)

7.742 = T_f - 20

T_f = 27.742

T_f = 28 °C