Question

A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is attached to a spring that has a spring constant k k of 205 N/m. 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass m w mw of the wooden block.

Answer 1

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

$mu=(M+m)v$ (1)

where

$m=11.9 g = 11.9\cdot 10^{-3}kg$ is the mass of the bullet

$u=261 m/s$ is the initial velocity of the bullet

$M$ is the mass of the block

$v$ is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

$\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2$ (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

$v=\frac{mu}{M+m}$

And substituting into eq(2), we can solve to find the mass of the block:

$(M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg$