Problemas
10. A .012 kg bullet with an initial velocity of 410m/s lodges into a piece of week and comes to rest within a distance of .085 m. What is the Impulse from the block on the bullet? 11. Cart A and cart B are on a flat tabletop with negligible friction Cart A moves toward Cart B at 1.70m/s , and Cart B is stationary. Cart A and Cart B has a mass of 3.0 kg After Cart A collides with Cart B, Cart A moves off with a velocity of .95m/s . What is Cart B's velocity after the collision?
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Ángela
professionell · Tutor durante 6 años
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Respuesta
10. The impulse experienced by the bullet from the block can be calculated using the formula:Impulse = Change in momentumThe initial momentum of the bullet is given by:Initial momentum = mass of bullet * initial velocity of bullet = 0.012 kg * 410 m/s = 4.92 kg*m/sSince the bullet comes to rest, its final momentum is 0. Therefore, the change in momentum is:Change in momentum = Final momentum - Initial momentum = 0 - 4.92 kg*m/s = -4.92 kg*m/sSince impulse is equal to the change in momentum, the impulse experienced by the bullet from the block is -4.92 kg*m/s.11. To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision.Let's denote the velocity of Cart B after the collision as v_B.The total momentum before the collision is:Total momentum before = mass of Cart A * velocity of Cart A = 3.0 kg * 1.70 m/s = 5.1 kg*m/sThe total momentum after the collision is:Total momentum after = mass of Cart A * velocity of Cart A after collision + mass of Cart B * velocity of Cart B after collision = 3.0 kg * 0.95 m/s + 3.0 kg * v_BSince the total momentum before and after the collision must be equal, we can set up the equation:5.1 kg*m/s = 3.0 kg * 0.95 m/s + 3.0 kg * v_BSolving for v_B, we get:v_B = (5.1 kg*m/s - 3.0 kg * 0.95 m/s) / 3.0 kg = 0.85 m/sTherefore, the velocity of Cart B after the collision is 0.85 m/s.