A block of mass M = 4.80 kg, at rest on a horizontal frictionless table, is atta

Question

A block of mass M = 4.80 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 5210 N/m. A bullet of mass m = 9.20 g and velocity ModifyingAbove v With right-arrow of magnitude 670 m/s strikes and is embedded in the block (the figure). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

A block of mass M = 4.80 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 5210 N/m. A bullet of mass m = 9.20 g and velocity ModifyingAbove v With right-arrow of magnitude 670 m/s strikes and is embedded in the block (the figure). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

A block of mass M = 4.80 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 5210 N/m. A bullet of mass m = 9.20 g and velocity ModifyingAbove v With right-arrow of magnitude 670 m/s strikes and is embedded in the block (the figure). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

Explanation / Answer

here,

initial speed of bullet , u = 670 m/s

M = 4.8 kg

m = 0.0092 kg

a)

let the speed of bullet immediately after the collison be v

using conservation of mommentum

m * u = ( M + m) * v

0.0092 * 670 = ( 4.8 + 0.0092 ) * v

v = 1.28 m/s

b)

let the amplitude of motion be A

using conservation of energy

0.5 * K * A^2 = 0.5 *(m + M) * v^2

5210 * A^2 = (4.8 + 0.0092) * 1.28^2

A = 0.04 m

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