Question

Two persons (each on a skateboard) are initially standing still and facing each other before pushing off of each other. One person (50 kg) moves backwards at 1.1 m/s. Calculate the velocity of the second person (65 kg).

Answer 1

Answer:

The second person will have a velocity of 0.85 m/s in the opposite direction of the first person.

Explanation:

Law Of Conservation Of Linear Momentum

The total momentum of a system of bodies is conserved unless an external force is applied. The formula for the momentum of a body with mass m and velocity v is:

P=mv.

If we have a system of bodies, then the total momentum is the sum of them all

$P=m_1v_1+m_2v_2+...+m_nv_n$

If a collision occurs, the velocities change to v' and the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

In a system of two masses, the law of conservation of linear momentum is:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

According to the conditions of the problem, two persons standing on a frictionless surface are initially at rest (v1=0, v2=0). Their masses are m1=50 Kg and m2=65 Kg. After the push, one person (say m1) moves backward at v1'=-1.1 m/s. We can calculate the speed of the other person by solving for v2':

$\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2)$

Substituting:

$\displaystyle v'_2=\frac{50*0+65*0-50(-1.1)}{65)$

$\displaystyle v'_2=\frac{55}{65}$

$v'_2=0.85 \ m/s$

The second person will have a velocity of 0.85 m/s in the opposite direction of the first person.