Bob, who has a mass of 80kg , can throw a 550g rock with a speed of 27m/s . The distance through which his hand moves as he accelerates the rock forward from rest until he releases it is 1.0 m.
What constant force must Bob exert on the rock to throw it with this speed?
If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?
To find the constant force that Bob must exert on the rock, we can use the principle of conservation of momentum. The initial momentum of the rock and Bob together is zero because they were at rest. After Bob throws the rock, both Bob and the rock move in opposite directions with their respective momenta.
1. Calculate the initial momentum:
Initial momentum = mass of rock * velocity of rock + mass of Bob * velocity of Bob
Initial momentum = (0.55 kg) * (0 m/s) + (80 kg) * (0 m/s) (since they were both at rest)
Initial momentum = 0 kg*m/s
2. Calculate the final momentum:
Final momentum = mass of rock * velocity of rock + mass of Bob * velocity of Bob
Final momentum = (0.55 kg) * (27 m/s) + (80 kg) * (-recoil velocity of Bob)
Final momentum = (0.55 kg) * (27 m/s) - (80 kg) * (recoil velocity of Bob)
According to the principle of conservation of momentum, the initial momentum and the final momentum should be equal:
0 kg*m/s = (0.55 kg) * (27 m/s) - (80 kg) * (recoil velocity of Bob)
3. Solve for the recoil velocity of Bob:
(-0.55 kg * 27 m/s) / 80 kg = recoil velocity of Bob
Now, let's calculate the recoil velocity of Bob:
recoil velocity of Bob = (-0.55 kg * 27 m/s) / 80 kg
recoil velocity of Bob = -0.1864 m/s
Therefore, the recoil speed of Bob after releasing the rock on the frictionless ice is approximately 0.1864 m/s (in the opposite direction).
Note: The negative sign indicates that Bob moves in the opposite direction to the rock.