Question

: A 10 g bullet travelling at 300 m/s hits a 500 g wooden block that is initially stationary. The bullet becomes embedded in the block, and both travel together along a frictioneless surface. Please answer each of the following questions a) What is the kinetic energy of the bullet before it hits the block? b) What is the velocity of the bullet+block after the collision?

Answer 1

Answer:

450 J

5.88235 m/s

Explanation:

The kinetic energy is given by

$K=\frac{1}{2}m_1u_1^2\\\Rightarrow K=\frac{1}{2}\times 0.01\times 300^2\\\Rightarrow K=450\ J$

The kinetic energy of the bullet before it hits the block is 450 J

$m_1$ = Mass of bullet = 0.01 kg

$m_2$ = Mass of block = 0.5 kg

$u_1$ = Initial Velocity of bullet = 300 m/s

$u_2$ = Initial Velocity of second block = 0 m/s

v = Velocity of combined mass

In this system the linear momentum is conserved

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{0.01\times 300 + 0.5\times 0}{0.01 + 0.5}\\\Rightarrow v=5.88235\ m/s$

The velocity of the bullet+block after the collision is 5.88235 m/s