Answer:
According to the law of conservation of momentum, the total momentum of the system before the gunshot is equal to the total momentum of the system after the deer scatter. Since the rock is initially at rest, the total momentum of the system before the gunshot is zero. Therefore, the total momentum of the system after the deer scatter must also be zero.
Let's add up the momenta of the three deer after they scatter:
p⃗ total = m1v⃗ 1 + m2v⃗ 2 + m3v⃗ 3
where m1 = m2 = m3 = 91.1 kg are the masses of the deer, and v⃗ 1, v⃗ 2, and v⃗ 3 are their respective velocities.
p⃗ total = (91.1 kg)(13.5 m/s)i^ + (91.1 kg)(4.2 m/s)j^
+ (91.1 kg)(-12.6 m/s)i^ + (91.1 kg)(9.1 m/s)j^
+ (91.1 kg)(2.1 m/s)i^ + (91.1 kg)(-17.7 m/s)j^
p⃗ total = (0 kgm/s)i^ + (0 kgm/s)j^
Since the total momentum of the system after the deer scatter is zero, the momentum of the rock must be equal in magnitude and opposite in direction to the momentum of the deer:
p⃗ rock = - p⃗ total
The mass of the rock and the three deer is 3(91.1 kg) + 274 kg = 547.3 kg. Therefore, the velocity of the rock is:
v⃗ rock = p⃗ rock / m_total
v⃗ rock = - p⃗ total / (3m + M)
v⃗ rock = - (0 kgm/s)i^ - (0 kgm/s)j^ / (3(91.1 kg) + 274 kg)
v⃗ rock = - (0 kgm/s)i^ - (0 kgm/s)j^ / 547.3 kg
v⃗ rock = 0 m/s i^ - 0 m/s j^
Therefore, the speed of the rock after the deer scatter is zero, which means that it remains at rest