1. Question The ways that mechanical energy is lost from the system in this expe

Question

1. Question The ways that mechanical energy is lost from the system in this experiment include: (Select all that apply.)

2. The ballistic pendulum (see figure a) is a device used to measure the speed of a fast-moving projectile such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet is stopped by the block, and the entire system swings up to a height h. It is possible to obtain the initial speed of the bullet by measuring h and the two masses. As an example of the technique, assume that the mass of the bullet, m1, is 4.71 g, the mass of the pendulum, m2, is 1.002 kg, and h is 5.18 cm. Find the initial speed of the bullet, v1i.

Explanation / Answer

according to law of conservation of momentum ,     m1v1i + m2v2i = (m1 + m2)vf     (0.00471 kg)(v1i) + (1.002 kg)(0) = (1.002 kg + 0.00471 kg)vf     (0.00471 kg)(v1i) = (1.00671 kg)vf  .................... (1) according to law of conservation of energy ,   (1/2)(m1 + m2)vf2 + 0 = 0 + (m1 + m2)gh          vf  = 2gh              = {(2)(9.8 m/s2)(5.18*10-2 m)}              = 1.007 m/s substitute this value in eq (1) , we get      (0.00471 kg)(v1i) = (1.00671 kg)(1.007 m/s) initial speed vi = 215.36 m/s .................................................................... for bullet - block system : from law of conservation of energy ,     (1/2)(m1 + m2)v2 = (1/2)kx2 velocity v = {kx2/(m1 + m2)}                = {(650 N/m)(6.45*10-2 m)2/(0.00543 kg + 2.155 kg)}                = 1.25 m/s for bullet  : according to law of conservation of momentum ,     m1vbullet + m2v2i = (m1 + m2)v     m1vbullet + 0 = (m1 + m2)v     bullet speed vbullet = (m1 + m2)v / m1                               = {(0.00543 kg + 2.155 kg)(1.25 m/s)} / (0.00543 m/s)                        = 497.33 m/s     (0.00471 kg)(v1i) + (1.002 kg)(0) = (1.002 kg + 0.00471 kg)vf     (0.00471 kg)(v1i) = (1.00671 kg)vf  .................... (1) according to law of conservation of energy ,   (1/2)(m1 + m2)vf2 + 0 = 0 + (m1 + m2)gh          vf  = 2gh              = {(2)(9.8 m/s2)(5.18*10-2 m)}              = 1.007 m/s substitute this value in eq (1) , we get      (0.00471 kg)(v1i) = (1.00671 kg)(1.007 m/s) initial speed vi = 215.36 m/s .................................................................... for bullet - block system : from law of conservation of energy ,     (1/2)(m1 + m2)v2 = (1/2)kx2 velocity v = {kx2/(m1 + m2)}                = {(650 N/m)(6.45*10-2 m)2/(0.00543 kg + 2.155 kg)}                = 1.25 m/s for bullet  : according to law of conservation of momentum ,     m1vbullet + m2v2i = (m1 + m2)v     m1vbullet + 0 = (m1 + m2)v     bullet speed vbullet = (m1 + m2)v / m1                               = {(0.00543 kg + 2.155 kg)(1.25 m/s)} / (0.00543 m/s)                        = 497.33 m/s     (0.00471 kg)(v1i) = (1.00671 kg)vf  .................... (1) according to law of conservation of energy ,   (1/2)(m1 + m2)vf2 + 0 = 0 + (m1 + m2)gh          vf  = 2gh              = {(2)(9.8 m/s2)(5.18*10-2 m)}              = 1.007 m/s substitute this value in eq (1) , we get      (0.00471 kg)(v1i) = (1.00671 kg)(1.007 m/s) initial speed vi = 215.36 m/s .................................................................... for bullet - block system : from law of conservation of energy ,     (1/2)(m1 + m2)v2 = (1/2)kx2 velocity v = {kx2/(m1 + m2)}                = {(650 N/m)(6.45*10-2 m)2/(0.00543 kg + 2.155 kg)}                = 1.25 m/s for bullet  : according to law of conservation of momentum ,     m1vbullet + m2v2i = (m1 + m2)v     m1vbullet + 0 = (m1 + m2)v     bullet speed vbullet = (m1 + m2)v / m1                               = {(0.00543 kg + 2.155 kg)(1.25 m/s)} / (0.00543 m/s)                        = 497.33 m/s velocity v = {kx2/(m1 + m2)}                = {(650 N/m)(6.45*10-2 m)2/(0.00543 kg + 2.155 kg)}                = 1.25 m/s for bullet  : according to law of conservation of momentum ,     m1vbullet + m2v2i = (m1 + m2)v     m1vbullet + 0 = (m1 + m2)v     bullet speed vbullet = (m1 + m2)v / m1                               = {(0.00543 kg + 2.155 kg)(1.25 m/s)} / (0.00543 m/s)                        = 497.33 m/s for bullet  : according to law of conservation of momentum ,     m1vbullet + m2v2i = (m1 + m2)v     m1vbullet + 0 = (m1 + m2)v   according to law of conservation of momentum ,     m1vbullet + m2v2i = (m1 + m2)v   bullet speed vbullet = (m1 + m2)v / m1                               = {(0.00543 kg + 2.155 kg)(1.25 m/s)} / (0.00543 m/s)                        = 497.33 m/s                               = {(0.00543 kg + 2.155 kg)(1.25 m/s)} / (0.00543 m/s)                        = 497.33 m/s

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