Cart and running person

Problem

A $200$ kg cart travels at $2{,}0$ m/s. An $80$ kg person jumps onto it while running at $5{,}0$ m/s in the same direction. What is the final velocity of the system?

Solution

The person lands on the cart and the two carry on together: a perfectly inelastic collision. Total momentum is conserved (no horizontal external force).

	mass (kg)	$v$ before (m/s)	$v$ after (m/s)
cart	$200$	$2{,}0$	$v$
person	$80$	$5{,}0$	$v$

Conservation: $m_c v_c + m_p v_p = (m_c + m_p)\,v$ $v = \frac{m_c v_c + m_p v_p}{m_c + m_p} = \frac{200 \cdot 2{,}0 + 80 \cdot 5{,}0}{200 + 80}$ $v = \frac{400 + 400}{280} = \frac{800}{280} \approx 2{,}9\ \mathrm{m/s}$

Final result: $\ev{v \approx 2{,}9\ \mathrm{m/s}}$

Links

Topics: Quantità di moto e urti Concepts: Conservazione della quantità di moto · Urto anelastico Skills: Conservazione della quantità di moto Methods: Metodo delle due tabelle parallele