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

The magnitud of the velocity is

`8.46m/s`

and the direccion:

`-28.3` degrees from the horizontal.

Step-by-step 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)=(m_(b)v_(1b)-m_(b)v_(2b))/(m_(s))`

Substituting known values:

`v_(2s)=(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)|=√((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.