Answer:
v_f = 10.85 m/s
Explanation:
We will apply the law of conservation of momentum here:
[tex]m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\[/tex]
where,
m₁ = mass of roller skater = 47 kg
m₂ = mass of bag = 6 kg
v_1i = initial speed of roller skater = 12 m/s
v_2i = initial speed of the bag = 0 m/s
v_1f = final speed of the roller skater = ?
v_2f = final speed of the bag = ?
Both the bag and the skater will have same speed at the end because kater is carrying the bag:
v_1f = v_2f = v_f
Therefore, the equation will become:
[tex](47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\[/tex]
v_f = 10.85 m/s
