Question

A 50 kg girl rides on a 4.9 kg skateboard. The girl on the skateboard moves at 2.1 m/s. If the girl jumps off the skateboard backward with a velocity of 0.6 m/s, how fast does the skateboard roll away? ANSWER ASAP

Answer 1

Answer:

The skateboard rolls away at 29.7 m/s

Explanation:

Law Of Conservation Of Linear Momentum

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2$

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

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2\qquad\qquad[1]$

A girl of m1=50 kg rides an m2=4.9 kg skateboard and the common speed is v1=v2=2.1 m/s.

The girl jumps off the skateboard backward with a speed of v1'=-0.6 m/s (negative because it's opposite to the original direction). It's required to find the final speed of the skateboard. It will be calculated by solving for v2':

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

$\displaystyle v'_2=\frac{50*2.1+4.9*2.1-50*(-0.6)}{4.9}$

Calculating:

$\displaystyle v'_2=\frac{145.29}{4.9}=29.7\ m/s$

The skateboard rolls away at 29.7 m/s