Answer to Question #285574 in Mechanics | Relativity for AAA

Question

A block with mass M is at rest on a frictionless, horizontal surface. The block is attached to an ideal, massless spring that initially is in equilibrium.

A riffle bullet with mass m_p​ is fired with speed v_b​ into the block and embeds itself in the block. The impact compresses the spring over a distance D. The block and bullet oscillate in simple harmonic motion.

(a) Calculate the angular frequency ω of the oscillation.

(b) When time t=0 is defined as the instant of impact, the positive x-axis is pointing to the right, and x=0 is the initial position of the block before impact, give an expression x(t) for the time dependent position of the block and bullet. Assume that the impact itself takes no time.

Answer 1

(a) The angular frequency is

$$\omega=\sqrt{k/(M+m_b)}.$$

By conservation of momentum:

$$m_bv_b=(M+m_b)u→u=\dfrac{m_bv_b}{M+m_b}.$$

By conservation of energy:

$$\frac12(M+m_b)u^2=\dfrac12kD^2,\\\space\\ \frac12·\dfrac{(m_bv_b)^2}{M+m_b}=\dfrac12kD^2,\\\space\\ k=\dfrac{(m_bv_bD)^2}{M+m_b}.$$

The angular frequency:

$$\omega=\dfrac{m_bv_bD}{M+m_b}.$$

(b) The expression is

$$x(t)=D\sin\bigg[\dfrac{m_bv_bD}{M+m_b}t\bigg].$$