Question

A 20-g bullet is shot vertically into an 2.8-kg block. The block lifts upward 9 mm. The bullet penetrates the block and comes to rest in it in a time interval of 5 ms. Assume the force on the bullet is constant during penetration and that air resistance is negligible. What is the speed of the bullet just before the impact

Answer 1

The speed of the bullet just before the impact is 701 m/s

Step-by-step explanation:

Given;

mass of the bullet, m₁ = 20 g = 0.02 kg

mass of the block, m₂ = 2.8 kg

displacement of the block, d = 9 mm = 9 x 10⁻³ m

duration of motion of the bullet, t = 5 ms = 5 x 10⁻³ s

Apply the principle of conservation of energy;

The final kinetic energy of the bullet = maximum potential energy of the block

`(1)/(2) m_1v^2 = m_2gh\\\\v^2 = (2m_2gh)/(m_1) \\\\v= \sqrt{(2m_2gh)/(m_1) } \\\\v = \sqrt{(2 * 2.8 * 9.8 * (9* 10^(-3)))/(0.02) } \\\\v = 4.97 \ m/s`

Apply the principle of conservation of linear momentum, to determine the initial velocity of the bullet before the impact.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

u₁ is the initial velocity of the bullet

u₂ is the initial velocity of the block = 0

m₁u₁ + 0 = v(m₁ + m₂)

m₁u₁ = v(m₁ + m₂)

0.02u₁ = 4.97(2.8 + 0.02)

0.02u₁ = 14.02

u₁ = 14.02 / 0.02

u₁ = 701 m/s

Therefore, the speed of the bullet just before the impact is 701 m/s