Answer:
Explanation:
Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .
velocity of approach = 1.5 - 0 = 1.5
velocity of separation = v₁ + v₂
coefficient of restitution = velocity of separation / velocity of approach
.8 = v₁ + v₂ / 1.5
v₁ + v₂ = 1.2
applying law of conservation of momentum
m x 1.5 + 0 = mv₂ - mv₁
1.5 = v₂ - v₁
adding two equation
2 v ₂= 2.7
v₂ = 1.35 m /s
v₁ = - .15 m / s
During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.
For second collision ,
coefficient of restitution = velocity of separation / velocity of approach
.5 = v₃ + v₄ / 1.35
v₃ + v₄ = .675
applying law of conservation of momentum
m x 1.35 + 0 = mv₄ - mv₃
1.35 = v₄ - v₃
adding two equation
2 v ₄= 2.025
v₄ = 1.0125 m /s
v₃ = - 0 .3375 m / s
H2O is water
CO2 is carbon dioxide
Answer:
156watts
Explanation:
M = 20
U = 0.200
Vo = 8.0
We are to get p
P = w/t
W = 1/2(vf²-Vo²)
The final velocity is 0
W =-1/2*20*8²
= -640J
Acceleration = -ug
= -0.200*9.8
= -1.96m/s^-2
We are to get t
t = 8/1.96
= 4.1s
P = w/t
= 640/4.1
= 156 watts
156watts is the average power that is produced by friction as the rock stops.
