Answer:
The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.
See attachment below for full solution
Explanation:
This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.
Answer:
v₂ = 5.7 m/s
Explanation:
We will apply the law of conservation of momentum here:

where,
Total Initial Momentum = 340 kg.m/s
m₁ = mass of bike
v₁ = final speed of bike = 0 m/s
m₂ = mass of Sheila = 60 kg
v₂ = final speed of Sheila = ?
Therefore,

<u>v₂ = 5.7 m/s </u>
Y, bc the height of the bounce back is higher than x