Answer:
The net force is 1350 kN
Solution:
As per the question:
Mass of man, m = 60 kg
Initial speed of the car, v = 15 m/s
Final speed of the car, v' = 0 m/s
Distance covered by the person before coming to rest, d = 5 mm =
By using the third eqn of motion:
Thus the force on the person can be given by:
F = ma =
Answer:
The velocity of skaters after collision is 0.55 m/s and they are moving to the left
Explanation:
Consider m₁ and m₂ are the masses of two skaters and their initial velocity be v₁ and v₂ respectively.
Consider the velocity along right direction as positive while negative for left direction.
According to the problem,
Mass of first skater, m₁ = 66.0 kg
Mass of second skater, m₂ = 75.0 kg
Velocity of first skater, v₁ = + 2.00 m/s
Velocity of second skater, v₂ = -2.80 m/s
Since, the two skaters grab each other after collision. Hence, they are moving with same final velocity v.
Applying conservation of linear momentum,
Momentum before collision = Momentum after collision
m₁v₁ + m₂v₂ = (m₁+m₂)v
Substitute the suitable values in the above equation.
66 x 2 - 75 x 2.8 = (66 + 75 )v
v = -0.55 m/s
The negative sign denotes that both the skaters are moving in the left direction.