A rifle bullet with mass 90 grams strikes and embeds itself

A rifle bullet with mass 9.0 grams strikes and embeds itself into a block with mass 992 grams that rests on a frictionless, horizontal surface. The block is attached to a spring. The impact causes the spring to compress 15.0 cm before stopping. A previous calibration of the spring found that it takes a force of 0.750 Newtons to compress the spring 0.25 cm Find the magnitude of the block\'s velocity immediately after impact What was the initial speed of the bullet? What was the impulse of the bullet on the block took 0.05 seconds for the bullet to embed itself into block?

Solution

We will use the principle of conservation of energy to determine the speed of the block after the collision using the maxium compression we get. Also, we will then use the principle of conservation of momentum to determine the speed of the bullet.

Part a.) It takes 0.75 N to compress the spring by 25 cm. So we have the spring constant as 3 N/m. The spring gets compressed by 15 cm and the mass of the block and the bullet together 1.001 Kg

So we can write:

1.001 x V² = 3 x 0.0225

or, V = 0.2597 m/s

So the velocity of the block immediately after the collision is 0.2597 m/s

Part b.) Now, we will use the principal of conservation of linear momentum here, to write:

mv = (M + m)V where m is the mass of the bullet and v is its velocity while (M + m) is the mass of the block and the bullet and V is the velocity after the collision.
So, we can write: 9 x v = 1001 x 0.2597

or, v = 28.882 m/s is the required speed of the bullet.

Part c.) Impulse as defined is the change in the momentum of the body.

So the impulse on the block by the bullet would be given as change in momentum of the block after the collision which would be given as Mass of block x velocity = 0.992 x 0.2597 = 0.2576 Kg-m/s