A 91 kg man lying on a surface of negligible friction shoves a 75 g stone away from himself, giving it a speed of 4.0 m/s. What speed does the man acquire as a result?
To find the speed the man acquires as a result of shoving the stone, we can use the principle of conservation of momentum. According to this principle, the total momentum before the shove is equal to the total momentum after the shove.
Before the shove:
The man's initial momentum is given by the equation:
momentum(man) = mass(man) x velocity(man)
The mass of the man is 91 kg, and since he is initially at rest, his velocity is 0 m/s. Therefore, momentum(man) = 91 kg x 0 m/s = 0 kg·m/s.
Since the stone is initially at rest, its momentum is also 0 kg·m/s: momentum(stone) = mass(stone) x velocity(stone) = 0.075 kg x 0 m/s = 0 kg·m/s.
Thus, the total momentum before the shove is 0 kg·m/s.
After the shove:
Let's define the speed of the man after the shove as v(man) and the speed of the stone after the shove as v(stone).
The momentum of the man after the shove is equal to the product of his mass and his velocity: momentum(man) = mass(man) x velocity(man) = 91 kg x v(man) kg·m/s.
Similarly, the momentum of the stone after the shove is given by: momentum(stone) = mass(stone) x velocity(stone) = 0.075 kg x 4.0 m/s = 0.3 kg·m/s.
Since momentum is conserved, the total momentum after the shove is 0 kg·m/s.
Therefore, we can write the equation:
momentum(man) + momentum(stone) = 0 kg·m/s + 0.3 kg·m/s
91 kg x v(man) kg·m/s + 0.3 kg·m/s = 0 kg·m/s + 0.3 kg·m/s
91 kg x v(man) kg·m/s = 0 kg·m/s
We can divide both sides of the equation by 91 kg to isolate v(man):
v(man) kg·m/s = 0 kg·m/s / 91 kg
v(man) = 0 m/s
Therefore, the man does not acquire any speed as a result of shoving the stone.