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

The skateboard rolls away at 29.7 m/s

Step-by-step 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=(m_1v_1+m_2v_2-m_1v'_1)/(m_2)`

`\displaystyle v'_2=(50*2.1+4.9*2.1-50*(-0.6))/(4.9)`

Calculating:

`\displaystyle v'_2=(145.29)/(4.9)=29.7\ m/s`

The skateboard rolls away at 29.7 m/s