A 70.0 kg football player is gliding across very smooth

Step 1 of 2

a) We need to find out the player's speed if the ball's velocity is \(15 \mathrm{~m} / \mathrm{s}\) relative to the ground.

Using the conservation of momentum we can write,

                                            \(m_{P}\left(v_{f}\right)_{P}+m_{B}\left(v_{f}\right)_{B}=m_{B} v_{i}+m_{P} v_{i}\)

                              \(m_{P}\left(v_{f}\right)_{P}+m_{B}\left(v_{f}\right)_{B}=\left(m_{B}+m_{P}\right) v_{i}\)

That is,

                                         \(m_{P}\left(v_{f}\right)_{P}=\left(m_{B}+m_{P}\right) v_{i}-m_{B}\left(v_{f}\right)_{B}\)

Implies,

                                         \(\left(v_{f}\right)_{P}=\frac{\left(m_{B}+m_{P}\right) v_{i}-m_{B}\left(v_{f}\right)_{B}}{m_{P}}\)

Mass of the ball is, \(m_{B}=0.450 \mathrm{~kg}\)

Mass of the person, \(m_{P}=70 \mathrm{~kg}\)

Final velocity of the ball, \(\left(v_{f}\right)_{B}=15 \mathrm{~m} / \mathrm{s}\)

Initial velocity of the person is equal to the initial velocity of the ball, \(v_{i}=2 \mathrm{~m} / \mathrm{s}\)

So the final velocity of the person,

                                              \(\left(v_{f}\right)_{P} =\frac{(0.450 \mathrm{~kg}+70 \mathrm{~kg})(2 \mathrm{~m} / \mathrm{s})-(0.450 \mathrm{~kg})(15 \mathrm{~m} / \mathrm{s})}{70 \mathrm{~kg}}\)

                                            \(\left(v_{f}\right)_{P} =1.92 \mathrm{~m} / \mathrm{s}\)

The final velocity of the person is \(1.92 \mathrm{~m} / \mathrm{s}\)