An (63Kg) astronaut, floating alone in outer space, catches a 7Kg bowling ball traveling at 20m/s. The velocity of the astronaut and bowling ball after the catch is
To find the velocity of the astronaut and the bowling ball after the catch, we can use the law of conservation of momentum.
The law of conservation of momentum states that the total momentum of a system remains constant if no external forces act on it. In this case, the astronaut and the bowling ball are the two objects in the system.
The momentum of an object is the product of its mass and velocity. Mathematically, momentum (p) is given by the formula:
p = m * v
where p is momentum, m is mass, and v is velocity.
Given:
Mass of the astronaut (ma) = 63 kg
Mass of the bowling ball (mb) = 7 kg
Velocity of the bowling ball (vb) = 20 m/s
To find the velocity of the astronaut and the bowling ball after the catch, we need to calculate the momentum of the system before and after the catch and equate them.
Before the catch, the momentum of the system is the sum of the individual momenta of the astronaut and the bowling ball:
Initial momentum = ma * va + mb * vb
Since the astronaut is initially at rest (va = 0), we can simplify the equation:
Initial momentum = mb * vb
After the catch, the momentum of the system is the sum of the individual momenta of the astronaut and the bowling ball:
Final momentum = (ma + mb) * vf
Since we're looking for the final velocity (vf), we can rearrange the equation:
vf = Final momentum / (ma + mb)
Now, let's plug in the values:
Initial momentum = 7 kg * 20 m/s = 140 kg*m/s
Final momentum = (63 kg + 7 kg) * vf
Equating the initial and final momentum:
140 kg*m/s = 70 kg * vf
Simplifying the equation:
vf = 140 kg*m/s / 70 kg
vf = 2 m/s
Therefore, the velocity of the astronaut and bowling ball after the catch is 2 m/s.