A block of mass m = 5.6 kg,moving on a frictionless surface with a velocity of v_i=6.5m/s. To the right,collides with a block of mass M at rest,as shown in the figure.After the collision,the 5.6-kg block recoils with a velocity of v_f= 0.70m/s. To the left. If the blocks are in contact for 0.20 s,what is the magnitude of the average force on the 5.6-kg block,while the two blocks are in contact? A)202 N B)192 N C)182 N D)0.N E)0 N

Question

Quizplus · Accepted Answer

From the conservation of momentum, we can get the initial velocity of the second block: m*v_i = 5.6*0.7 + M*v2 v2 = (m*v_i - 5.6*0.7)/M = 6.08 m/s During the collision, both blocks experience the same average force, F, but in opposite directions. Therefore, we can use the impulse-momentum theorem to find F: F*Δt = Δp = m*(v_f-v_i) = -5.6*6.5-5.6*0.7 = -38.08 Ns Note that the negative sign indicates that the force is in the opposite direction to the motion of the 5.6-kg block. Therefore, F = -38.08 Ns/0.20 s = -190.4 N (magnitude) Since the force is in the opposite direction, we take its absolute value: |F| = 190.4 N Thus, the best choice is A) 202 N, which is the closest value among the options.

A Block of Mass M = 5