Answer:
v=32.49 m/s
Explanation:
Given that
Distance ,d= 66 m
Initial speed of the car ,u = 0 m/s
Coefficient of friction ,μ = 0.8
Lets take the total mass of the car = m
The acceleration of the car is given as
a = μ g ( g= 10 m/s² )
Now by putting the values in the above equation we get
a= 0.8 x 10 m/s²
a= 8 m/s²
We know that ,final speed is given as
v²= u ²+ 2 a d
Now putting the value
v²=0² + 2 x 8 x 66
v²= 1056
v=32.49 m/s
Answer
given,
mass of ball, m = 57.5 g = 0.0575 kg
velocity of ball northward,v = 26.7 m/s
mass of racket, M = 331 g = 0.331 Kg
velocity of the ball after collision,v' = 29.5 m/s
a) momentum of ball before collision
P₁ = m v
P₁ = 0.0575 x 26.7
P₁ = 1.535 kg.m/s
b) momentum of ball after collision
P₂ = m v'
P₂ = 0.0575 x (-29.5)
P₂ = -1.696 kg.m/s
c) change in momentum
Δ P = P₂ - P₁
Δ P = -1.696 -1.535
Δ P = -3.231 kg.m/s
d) using conservation of momentum
initial speed of racket = 0 m/s
M u + m v = Mu' + m v
M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)
0.331 u' = 3.232
u' = 9.76 m/s
change in velocity of the racket is equal to 9.76 m/s