A 39-kg girl standing at rest on slippery (frictionless) ice throws a 10-g ball forward with a speed of 10 m/s. Just afterward, the girl will
at rest
2.5m/s
To find out what will happen to the girl just after she throws the ball, we can use the principle of conservation of momentum.
The momentum of an object is given by the product of its mass and velocity. In this case, we have:
Girl's mass (m1) = 39 kg
Girl's velocity before throwing the ball (v1) = 0 m/s (since she is at rest)
Ball's mass (m2) = 10 g = 0.01 kg
Ball's velocity after being thrown (v2) = 10 m/s
To apply conservation of momentum, we need to consider the total momentum before and after the ball is thrown.
Total momentum before = m1 * v1 + m2 * v1 (since the girl is at rest, her velocity is 0)
Total momentum after = m1 * v3 + m2 * v2 (v3 is the girl's velocity just after she throws the ball)
Since we are assuming there is no external force acting on the system, the total momentum before and after must be the same. Therefore, we can write:
m1 * v1 + m2 * v1 = m1 * v3 + m2 * v2
Substituting the given values:
39 kg * 0 m/s + 0.01 kg * 0 m/s = 39 kg * v3 + 0.01 kg * 10 m/s
Simplifying the equation:
0 kg m/s = 39 kg * v3 + 0.1 kg m/s
Now, to find v3, we rearrange the equation:
39 kg * v3 = -0.1 kg m/s
Dividing both sides by 39 kg:
v3 = -0.1 kg m/s / 39 kg
Calculating the value:
v3 ≈ -0.0026 m/s
Therefore, just after the girl throws the ball, she will have a backward velocity of approximately -0.0026 m/s, which means she will move slightly backward on the slippery ice.