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 exerted 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 exerted 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 use conservation of momentum to find the velocity of the second block right after the collision:
\begin{align*}
m v_i &= m v_{1f} + M v_{2f} \
5.6 \cdot 6.5 &= 5.6 \cdot (-0.70) + Mv_{2f} \
Mv_{2f} &= 54.88 + 36.96 \
v_{2f} &= \frac{91.84}{M}
\end{align*}
Next, we can use the impulse-momentum theorem to find the average force:
\begin{align*}
F_{avg} &= \frac{\Delta p}{\Delta t} \
&= \frac{m\Delta v}{\Delta t} \
&= \frac{m(v_{1f} - v_i)}{\Delta t} \
&= \frac{5.6(-0.70 - 6.5)}{0.20} \
&= 202 	extrm{ N}
\end{align*}
Therefore, the answer is (A) 202 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 $v _ { \mathrm { i } } = 6.5 \mathrm {~m} / \mathrm { s }$