Question

A 1.14 kg block is attached to a horizontal spring with spring constant 1700 N/m . The block is at rest on a frictionless surface. A 11 g bullet is fired into the block, in the face opposite the spring, and sticks.What is the bullet's speed if the subsequent oscillations have an amplitude of 10.0 cm? (Express your answer in m/s as a pure number).

Answer 1

Answer:

402 m/s

Explanation:

$m_1$ = Mass of bullet = 11 g

$m_2$ = Mass of block = 1.14 kg

v = Velocity of bullet

u = Velocity of block and bullet

x = Displacement of spring = 10 cm

k = Spring constant = 1700 N/m

The momentum of the system is conserved

$m_1v=(m_1+m_2)u\\\Rightarrow 0.011v=(0.011+1.14)u\\\Rightarrow 0.011v=1.151u$

Also,

$\dfrac{1}{2}kx^2=\dfrac{1}{2}(m_1+m_2)u^2\\\Rightarrow u=\sqrt{\dfrac{kx^2}{m_1+m_2}}\\\Rightarrow u=\sqrt{\dfrac{1700\times 0.1^2}{1.14+0.011}}\\\Rightarrow u=3.84314\ m/s$

So,

$0.011v=1.151u\\\Rightarrow v=\dfrac{1.151u}{0.011}\\\Rightarrow v=\dfrac{1.151\times 3.84314}{0.011}\\\Rightarrow v=402.13219\ m/s$

Velocity of bullet is 402 m/s