Question

A 12.0 g bullet is fired horizontally into 100 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring constant of 150N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 80.0 cm, what was the speed of the bullet at impact with the block?

Answer 1

From the law of conservation of momentum:

m_1 v=(m_1+m_2 ) v_0

and from the law of conservation of energy:


((m_1+m_2 ) v_0^2)/2=(kx^2)/2

where m_1 is mass of the bullet; m_2 is mass of the block; k is a spring constant; x is a maximum of spring compressing; v is the speed of bullet at impact with the block; v_0 is the speed of the bullet-block system just after the impact.
Then:

v=(m_1+m_2)/m_1 v_0

v_0=x√(k/(m_1+m_2 ))

v=x/m_1 √(k(m_1+m_2 ) )

Answer:

v=0.8/0.012 √(150∙(0.012+0.1) )=273 m/s