A 4 kg wooden block renting on a horizontal surface with a coefficient of fricti

Question

A 4 kg wooden block renting on a horizontal surface with a coefficient of friction 0.4 is struck by a bullet of maze 5 gram traveling horizontally at 700 n/e. The bullet retrains embedded In the block after the collision. Immediately after the collision. the momentum of the system is kgn/s. 3.5 8.75 2800 1120 1225 Immediately after the collision, the kinetic energy of the block in -- J. 2600 1225 875 1.53 The block will slide m before coming to rest. 0.052 0.098 0.175 0.392 A mass of 25 kg is suspended by a light string from a pulley of radius 0.25 m and moment of inertia 0.8 kgm^2. The. pulley turns on bearings of negligible friction. When the mass descends 2 m the pulley turns through rad. 2 4 6 0 10 When the mass descends through 2 m, the potential energy of the mass decreases by J 0 119 392 425 490 The acceleration of the mass is m/s^2 A. 5.43 6-48 7-42 8.22 9.8

Explanation / Answer

14. Momentum of bullet plus block system will be constant for collision,

hence momentum = (0.005 * 700) + (4 * 0) = 3.5 kg m/s

15. after collision block and bullet will move with same veocity.

hence momentum = (m + M)v

3.5 = (4 + 0.005)v

v = 0.87 m/s

KE = 4 * 0.87^2 /2 = 1.53 J

16. friction force = uk N = uk m g

using work energy theorem,

work done by friction = change in KE

- uk m g d = 0 - 1.53

d = 1.53 / (0.4 * 4 * 9.8) = 0.098 m

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