Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg.
1 answer:
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Answer:
u = 4.6 m/s
h = 8.01 m
Explanation:
Given:
Mass of the tennis ball, m = 44.0 g
Mass of the basket ball, M = 594 g
Height of fall, h = 1.08m
Now,
we have
where, s = distance = h
a = acceleration
u = final speed before the collision
u' = initial speed
since it is free fall case
thus,
a = g = acceleration due to gravity
u' = 0
thus we have
or
or
u = 4.6 m/s
b) Now after the bounce, the ball moves with the same velocity
thus, v = v₂
thus,
final speed () = v = 4.6 m/s
Then conservation of energy says
also
applying the concept of conservation of momentum
we have
mu₁ + Mu₂ = mv₁ + Mv₂
u₁ =velocity of the tennis ball before collision = -4.6 m/s
u₂ = velocity of the basketball before collision= 4.6 m/s
v₁ = velocity of the tennis ball after collision
v₂ = velocity of the basketball after collision
substituting the values in the equation, we get
Now,
solving both the equations simultaneously we get
substituting the values in the above equation we get
or
or
here negative sign depicts the motion of the ball in the upward direction
now the kinetic energy of the tennis ball
or
or
K.E = 3.45 J
also at the height the K.E will be the potential energy of the tennis ball
thus,
3.45 J = mgh
or
3.45 = 44 × 10⁻³ × 9.8 × h
h = 8.01 m