Question

0.180-kg stone rets on a frictionless, horizontal surface. A bullet of mass 7.50 g, traveling horizontally at 390 m/s, strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 210 m/s.
(a) Compute the magnitude and direction of the velocity of the stone after it is struck.

Answer 1

Answer:

The magnitud of the velocity is

$8.46m/s$

and the direccion:

$-28.3$ degrees from the horizontal.

Explanation:

Fist we define our variables:

$m_{s}=0.18kg\\m_{b}=7.5g = 0.0075kg\\v_{1b}= 390m/s -i\\v_{1s}=0m/s\\v_{2b}=210m/s -j\\v_{2s}=?$

The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.

velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.

Using conservation of momentum:

$m_{s}v_{1s}+m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}\\v_{1s}=0, so\\m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}$

Clearing for the velocity of the stone after the crash:

$v_{2s}=\frac{m_{b}v_{1b}-m_{b}v_{2b}}{m_{s}}$

Substituting known values:

$v_{2s}=\frac{0.0075kg(((390m/s-i)-210m/s-j)}{0.18kg}\\v_{2s}=(16.25m/s-i) - (8.75m/s-j)$

The magnitud of the velocity is :

$|v_{2s}|=\sqrt{(16.25m/s-i)^2 + (8.75m/s-j)^2}\\|v_{2s}|=18.46m/s$

and the direction:

$tan^{-1}(-8.75/16.25)=-28.3$

this is -28.3 degrees from the +i direction or the horizontal direcction.

Note: i and j can also be seen as x and y axis.