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

Explanations & Calculations

• You can apply the theory of conservation of linear momentum between the block and the bullet during the impact to find out the speed (V) of the block-bullet combined unit.
• Since the spring was at rest on a horizontal plane, that point marks the centre of the simple harmonic motion the system would undergo after the impact.
• The maximum deflection from that point to either direction hence could be taken as the amplitude of the SHM and n this case it is "D".
• Using the basic formulae for SHM, those needed details can be found.

$\qquad\qquad \begin{aligned} \leftarrow\\ \small M\times0+m_p.v_b&=\small (M+m_p).V_s\\ \small V_s&=\small \frac{m_pv_b}{(M+m_p)}\\ \\ \small v^2&=\small \omega^2(A^2-x^2)\\ \small v&=\small A\omega\qquad\cdots\cdots\cdots[\text{as x = 0 at the centre}]\\ \small \omega &=\small \frac{m_pv_b}{D(M+m_p)}\\ \\ \small x(t)&=\small -A\sin(\omega t)\\ \small x(t) &=\small A\sin(\pi+\omega t)\\ &=\small D\sin\bigg[\pi+\frac{m_pv_bt}{D(M+m_p)}\bigg] \end{aligned}$