A block of mass m = 5.6 kg, moving on a frictionless surface with a velocity of $v _ { \mathrm { i } } = 6.5 \mathrm {~m} / \mathrm { s }$ 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 $v _ { \mathrm { f } } = 0.70 \mathrm {~m} / \mathrm { s }$ 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

Question

A block of mass m = 5.6 kg, moving on a frictionless surface with a velocity of $v _ { \mathrm { i } } = 6.5 \mathrm {~m} / \mathrm { s }$ 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 $v _ { \mathrm { f } } = 0.70 \mathrm {~m} / \mathrm { s }$ 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

Quizplus · Accepted Answer

First, we need to solve for the velocity of the second block after the collision using conservation of momentum:
$$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$$
$$5.6(6.5) + 0 = 5.6(-0.70) + M(v_{2f})$$
$$v_{2f} = \frac{5.6(6.5) + 5.6(0.70)}{M}$$
Next, we can solve for the average force on the 5.6-kg block using the impulse-momentum theorem:
$$F_{avg} \Delta t = \Delta p$$
$$F_{avg} (0.20) = (5.6)(-0.70 - 6.5)$$
$$F_{avg} = \frac{(5.6)(-0.70 - 6.5)}{0.20} \approx -202 	ext{ N}$$
Since the force is acting in the opposite direction of the initial velocity, the magnitude of the average force is $|-202 	ext{ N}| = \boxed{	extbf{(A) } 202 	ext{ N}}$.

A Block of Mass M = 5 vi=6.5 m/sv _ { \mathrm { i } } = 6.5 \mathrm {~m} / \mathrm { s }vi​=6.5 m/s

A Block of Mass M = 5 $v _ { \mathrm { i } } = 6.5 \mathrm {~m} / \mathrm { s }$