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a 13-gram bullet, moving at 270 m/s, penetrates a 2 kg block of wood and emerges at a speed of 130 m/s. if teh block sits one a fricitonless surface, find its velocity after the bullet emerges.

Respuesta :

abidemiokin

Answer:

1.52m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute the given values into the formula

0.013(270)+2(130) = (270+130)v

3.51+260 = 400v

263.51 = 400v

v = 400/263.51

v = 1.52m/s

Hence the velocity after the bullet emerges is 1.52m/s

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