Question

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 2.8 m/s to the right towards a 37 kg girl who catches it. the girl is also on ice skates. what are the final speeds of the boy and the girl?

Answer 1

Let's split the problem into two parts, and let's solve it by using conservation of momentum.

1) Let's analyze the moment when the boy tosses the pumpkin. Before this moment, the boy is standing with the pumpkin, so with speed v=0 and therefore the total momentum of the system (boy+pumpkin) is zero.
After he tosses the pumpkin, the boy will start to move with speed $v_b$ and the pumpkin starts to move with speed $v_p=2.8 m/s$. The momentum of the boy is $m_b v_b$ (with $m_b =45 kg$ being the mass of the boy), while the momentum of the pumpkin is $m_p v_p$ (where $m_p=12 kg$ is the mass of the pumpkin). Since the total momentum must be equal to zero (because the total momentum cannot change), then we can write
$m_b v_b + m_p v_p =0$
From which we find
$v_b = - \frac{m_p v_p}{m_b}=- \frac{(12 kg)(2.8 m/s)}{45 kg} =-0.75 m/s$
and this is the speed of the boy, and the negative sign means he's moving in the opposite direction of the pumpkin.

2) Now let's focus on the entire system (boy+pumpkin+girl). Initially, the total momentum of this system is zero, because both the boy (holding the pumpkin) and the girl are still. So, the total momentum after the girl catches the pumpkin must be still zero.
After this moment, the boy has a momentum of $m_b v_b$, while the girl has momentum $(m_g+m_p)v_g$ where $m_g=37 kg$ is the girl mass $v_g$ is the girl speed. Here we use $m_g+m_p$ because the girl is holding the pumpkin now. Therefore, the conservation of momentum becomes
$m_bv_b + (m_g+m_p)v_g =0$
and so
$v_g = - \frac{m_b v_b}{m_g+m_p} =- \frac{(45 kg)(-0.75 m/s)}{37 kg+12 kg} =0.69 m/s$
and this is the girl's speed, with positive sign so with same direction of the pumpkin initial direction.