# A 45kg boy is holding a 12kg pumpkin while standing on ice skates on a smooth frozen pond. The boy tosses the pumpkin with a horizontal speed of 2.8m/s to a 37kg girl (also wearing ice skates) catches it. What are the final speeds of the boy and girl?

Apr 8, 2018

We need to use the Law of Conservation of Momentum.

1. For the boy standing and holding the pumpkin. Initial momentum $= 0$. In the absence of external force, after tossing the pumpkin total momentum of Boy+Pumpkin system must also be $= 0$.
Let ${v}_{b}$ be velocity of boy after tossing. Setting up the equation we get

$12 \times 2.8 + 45 \times {v}_{b} = 0$
$\implies {v}_{b} = - \frac{33.6}{45}$
$\implies {v}_{b} = - 0.74 \overline{6} \setminus m {s}^{-} 1$
Boy's final speed $| {v}_{b} | = 0.75 \setminus m {s}^{-} 1$

2. For the girl+pumpkin system. Initial momentum is of pumpkin only. After catching the pumpkin, girl+pumpkin system will have momentum equal to the initial momentum. Let ${v}_{g}$ be velocity of girl after catching.Therefore,

$12 \times 2.8 = \left(37 + 12\right) \times {v}_{g}$
$\implies {v}_{g} = \frac{33.6}{49}$
$\implies {v}_{g} = 0.69 \setminus m {s}^{-} 1$