It could be the third and the first one.
The correct answer to the question is : C) The horizontal momentum and the vertical momentum are both conserved.
EXPLANATION :
Before coming into any conclusion, first we have to understand the law of conservation of momentum.
As per the law of conservation of momentum, the total linear as well as angular momentum of an isolated system is always conserved . The law of conservation of energy is a universal fact.
Hence, during any type of collision, the total momentum is always conserved.
Hence, the total horizontal momentum as well as total vertical momentum are always conserved during both elastic as well as inelastic collision.
Answer:
D. 12.4 m
Explanation:
Given that,
The initial velocity of the ball, u = 18 m/s
The angle at which the ball is projected, θ = 60°
The maximum height of the ball is given by the formula
h = u² sin²θ/2g m
Where,
g - acceleration due to gravity. (9.8 m/s)
Substituting the values in the above equation
h = 18² · sin²60 / 2 x 9.8
= 18² x 0.75 / 2 x 9.8
= 12.4 m
Hence, the maximum height of the ball attained, h = 12.4 m
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