A block with mass M = 3 kg is moving on a flat surface with constant speed v1 = 12 m/s. A bullet with mass m = 0,1 kg is shot at an angle α = 20◦ with respect to the direction of the movement of the block. The speed of the bullet is v2 = 400 m/s. At what angle and with what speed does the block move after it is hit by the bullet that remains stuck inside the block? Draw a sketch. How long does it take for the block to stop moving if the friction coefficient between the block and the surface is kt = 0,2?

The speed does the block move after it is hit by the bullet that remains stuck inside the block will be 23.7 m/sec and it takes 12.07 seconds to stop.

What is the law of conservation of momentum?

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

Apply the law of conservation of momentum principle;

m₁v₁+m₂v₂cosΘ =(m₁+m₂)V

3 kg × 12 m/s + 0,1 kg × 400 m/s cos 20° = (3+0.1)V

36 + 40 cos 20° = 3.1 V

V=23.7 m/sec

The time it takes to stop when the friction coefficient between the block and the surface is 0.2 is found as;

V = u +at

V = 0+ μgt

23..7=0.2× 9.81 ×t

t=12.07 sec

Hence, it takes 12.07 seconds to stop.

To learn more about the law of conservation of momentum refer;

https://brainly.com/question/1113396

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