Answer:
The final velocity of the car A is -1.053 m/s.
Explanation:
For an elastic collision both the kinetic energy and the momentum of the system are conserved.
Let us call
= mass of car A;
= the initial velocity of car A;
= the final velocity of car A;
and
= mass of car B;
= the initial velocity of car B;
= the final velocity of car B.
Then, the law of conservation of momentum demands that
And the conservation of kinetic energy says that
These two equations are solved for final velocities and to give
by putting in the numerical values of the variables we get
and
Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.
Answer:
79.2 m/s
Explanation:
θ = angle at which projectile is launched = 29.7 deg
a = initial speed of launch = 130 m/s
Consider the motion along the vertical direction
v₀ = initial velocity along the vertical direction = a Sinθ = 130 Sin29.7 = 64.4 m/s
y = vertical displacement = - 108 m
a = acceleration = - 9.8 m/s²
v = final speed as it strikes the ground
Using the kinematics equation
v² = v₀² + 2 a y
v² = 64.4² + 2 (-9.8) (-108)
v = 79.2 m/s
Answer:
(a). The value of angular speed of the merry-go-round = - 5.82 ×
(b). The linear speed of the girl after the rock is thrown V = -1.89 ×
Explanation:
Given data
Mass of the girl = 50.6 kg
Mass of merry-go-round = 827 kg
Radius r = 3.72 m
The speed of the rock relative to the ground = 7.82
(a). The angular speed of the merry-go-round is given by
Put all the values in above formula
= - 5.82 ×
This is the value of angular speed of the merry-go-round.
(b). The liner speed of the girl is given by
⇒ V = r ×
⇒ V = - 3.72 × 5.82 ×
⇒ V = -1.89 ×
This is the linear speed of the girl after the rock is thrown.