Question

Procedure I. As before, perform two different collisions involving different masses of your own choosing. For Trial 1, let the target be initially at rest. For Trial 2, both carts should have nonzero velocities. Complete Table 7 below. Table 7 After Collision Before Collision Target Projectile Target Projectile Trial 5o 0.4 55 2565 lo.5 5 o.39 0. 12 -0.385 2. Calculate the total momentum of the system just before and just after the collision for each trial. Complete the Table 8 on the next page. Total Momentum Before Collision Total Momentum After Collision Table 8 Trial Questions: How did the total momentum of the system just before the collision compare to the total momentum of the system just after the collision for each trial?

need table 8 completed and the question answered

Answer 1

Solution:

Given that:

Given that By the given table: We have, m_1 = 505 g m_2 = 255 g V_1 initial = -0.4 g m s^-1 V_1 final = -0.12 ms^-1 V_2 initial = 0 V_2 final = -0.58 ms^-1 Total kinetic energy before collision = 1/2 m_1v_1^2 initial + 1/2 m_2 V_2^2 initial = 1/2 (505) (0.49)^2 + 1/2 times 255 times (0)^2 = 6.62 + 0 = 60.62 Joules \r\n Total kinetic energy after collision, = 1/2 m_1 v_1^2 final + 1/2 m_2 v_2^2 final = 1/2 times 505 times (-0.12)^2 + 1/2 times 255 times (0.58)^2 = 3.636 + 42.9 = 46.53 joules For trail 1 KE = KE by the given table trial two we have, M_1 = 505 g m_2 255 g V_initial = 0.25 ms^-2 V_2 initial = 0.318 ms^-1 V_1 final = 0.12 ms^-2 V_2 final = 0.385 ms^-1 Total kinetic energy before collision = 1/2 times 505