An 8.0-g bullet is suddenly shot into a 4.0-kg block, at rest on a frictionless

Question

An 8.0-g bullet is suddenly shot into a 4.0-kg block, at rest on a frictionless horizontal surface, as shown in the figure. The bullet remains lodged in the block. The block then moves against a spring and compresses it by 8.9 cm. The force constant (spring constant) of the spring is 1400 N/m. What is the magnitude of the impulse on the block (including the bullet inside) due to the spring during the entire time interval during which the block is in contact with the spring?

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6.7 N • s , 8.3 N • s, 10 N • s, 12 N • s, 13 N • s

Explanation / Answer

using law of conservation of energy

energy before collision =energy after collision

0.5*m*V^2 = 0.5*k*x^2

0.5**4.008*V^2 = 0.5*1400*0.089^2

V = 1.66 m/sec is the speed of the block and bullet before hitting the spring

then impulse is J = change in momentum

J = m*(V-u) = 4.008*1.66 = 6.7 Ns

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