Question

Two bodies of masses 5 and 7 kilograms are initially at rest on a horizontal frictionldess surface. A light spring is compressed between the bodies, which are held together by a thin thread. After the spring is released by burning through the thread, the 5 kilogram body has a speed of 0.2 m/s. The speed of the 7 kilogram body is ________ (in m/s).

Answer 1

Answer:

The veolcity of the 7 kilogram body is approximately 0.14 meters per second

Explanation:

Here we have to use the conservation of linear momentum for the bodies asumming the force of the spring an internal force on the system formed by the two blocks. So the change on linear momentum ($\overrightarrow{p}$) has to be zero.

$\varDelta\overrightarrow{p}=0$

$\overrightarrow{p_{f}}=\overrightarrow{p_{i}}$

The final momentum $\overrightarrow{p_{f}}$ is the sum of the final momentum of each block, and initial linear momentum $\overrightarrow{p_{i}}$ is zero because the blocks were initially at rest, so:

$m_{5}\overrightarrow{v_{5}}+m_{7}\overrightarrow{v_{7}}=0$

It's important to note that the linear momentum is a vector quantity so the direction of the velocities are important, assuming positive the velocity of the 5 kilogram body

$m_{5}v_{5}=-m_{7}v_{7}$

$v_{7}=\frac{-m_{5}v_{5}}{m_{7}}=\frac{-(5)(0.2)}{7}$

$v_{7}\approx-0.14\,\frac{m}{s}$

The speed is 0.14 meters per second and the negative sign means the direction is opposite of the 5 kg body velocity.