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?

10. The impulse experienced by the bullet from the block can be calculated using the formula:<br /><br />Impulse = Change in momentum<br /><br />The initial momentum of the bullet is given by:<br /><br />Initial momentum = mass of bullet * initial velocity of bullet<br /> = 0.012 kg * 410 m/s<br /> = 4.92 kg*m/s<br /><br />Since the bullet comes to rest, its final momentum is 0. Therefore, the change in momentum is:<br /><br />Change in momentum = Final momentum - Initial momentum<br /> = 0 - 4.92 kg*m/s<br /> = -4.92 kg*m/s<br /><br />Since impulse is equal to the change in momentum, the impulse experienced by the bullet from the block is -4.92 kg*m/s.<br /><br />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.<br /><br />Let's denote the velocity of Cart B after the collision as v_B.<br /><br />The total momentum before the collision is:<br /><br />Total momentum before = mass of Cart A * velocity of Cart A<br /> = 3.0 kg * 1.70 m/s<br /> = 5.1 kg*m/s<br /><br />The total momentum after the collision is:<br /><br />Total momentum after = mass of Cart A * velocity of Cart A after collision + mass of Cart B * velocity of Cart B after collision<br /> = 3.0 kg * 0.95 m/s + 3.0 kg * v_B<br /><br />Since the total momentum before and after the collision must be equal, we can set up the equation:<br /><br />5.1 kg*m/s = 3.0 kg * 0.95 m/s + 3.0 kg * v_B<br /><br />Solving for v_B, we get:<br /><br />v_B = (5.1 kg*m/s - 3.0 kg * 0.95 m/s) / 3.0 kg<br /> = 0.85 m/s<br /><br />Therefore, the velocity of Cart B after the collision is 0.85 m/s.