Question

Two blocks of masses mA and mB are connected by a massless spring. The blocks are moved apart, stretching the spring, and subsequently released from rest. Find (a) the ratio of velocities 7of the blocks at any point of their ensuing motion (when their velocities are non-zero) and (b) the ratio of the kinetic energies of the blocks.

Answer 1

Answer:

Part a)

$\frac{v_A}{v_B} = -\frac{m_B}{m_A}$

Part b)

$\frac{K_A}{K_B} = \frac{m_B}{m_A}$

Explanation:

Part a)

As we know that initially the two blocks are connected by a spring and initially stretched by some amount

Since the two blocks are at rest initially so its initial momentum is zero

since there is no external force on this system so final momentum is also zero

$m_Av_{1i} + m_Bv_{2i} = m_Av_A + m_Bv_B$

now for initial position the speed is zero

$0 = m_Av_A + m_Bv_B$

now we have

$\frac{v_A}{v_B} = -\frac{m_B}{m_A}$

Part b)

now for ratio of kinetic energy we know that the relation between kinetic energy and momentum is given as

$K = \frac{P^2}{2m}$

now for the ratio of energy we have

$\frac{K_A}{K_B} = \frac{P^2/2m_A}{P^2/2m_B}$

since we know that momentum of two blocks are equal in magnitude so we have

now we have

$\frac{K_A}{K_B} = \frac{m_B}{m_A}$