A thin block of soft wood with a mass of 0.086 kg rests on a horizontal friction

Question

A thin block of soft wood with a mass of 0.086 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 597 m/s at a block of wood and passes completely through it. The speed of the block is 21 m/s immediately after the bullet exits the block.
(a) Determine the speed of the bullet as it exits the block.
(b) Calculate the initial and final kinetic energies of the system.
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As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. The bullet has a mass m, a speed v before the collision with the target, and a speed
(0.536)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.433)KEb BC] of the kinetic energy of the bullet before the collision. Determine the mass M of the target and the speed V of the target the instant after the collision in terms of the mass m of the bullet and speed v of the bullet before the collision. (Express your answers to at least 3 decimals.)
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An object with a mass of 6.00 g is moving to the right at 14.0 cm/s when it is overtaken by an object with a mass of 26.0 g moving in the same direction with a speed of 22.0 cm/s. If the collision is elastic, determine the speed of each object after the collision.
26.00 g object=___ cm/s
6.00 g =___ cm/s

Explanation / Answer

here,

mass of bullet , m = 0.00467 kg

mass of block , M = 0.086 kg

initial speed of bullet , u1 = 597 m/s

final speed of block , v2 = 21 m/s

let the final speed of bullet be v1

using conservation of angular momentum

m * u1 = m * v1 + M * v2

0.00467 * 597 = 0.00467 * v1 + 0.086 * 21

v1 = 210.3 m/s

a)

the speed of bullet as it exits the block is 210.3 m/s

b)

initial kinetic energy of the system , KEi = 0.5 * m * u1^2

KEi = 0.5 * 0.00467 * 597^2

KEi = 832.2 J

final kinetic energy of the system , KEf = 0.5 * m * v1^2 + 0.5 * M * v2^2

KEf = 0.5 * ( 0.00467 * 210.3^2 + 0.086 * 21^2)

KEf = 122.2 J

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