<u>Explanation:</u>
Velocity of B₁ = 4.3m/s
Velocity of B₂ = -4.3m/s
For perfectly elastic collision:, momentum is conserved
where,
m₁ = mass of Ball 1
m₂ = mass of Ball 2
v₁ = initial velocity of Ball 1
v₂ = initial velocity of ball 2
v'₁ = final velocity of ball 1
v'₂ = final velocity of ball 2
The final velocity of the balls after head on elastic collision would be
Substituting the velocities in the equation
If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.
Answer:
In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction
Explanation:
Answer:
the answers are the first and last option
Answer
given,
Weight of the child = 110 N
length of the swing,L = 2 m
now, calculating the potential energy when the string is horizontal
Potential energy = m g h
now, h = L (1 - cos θ) where θ is the angle made by the string with the vertical.
PE = m g L (1 - cos θ)
when rope is horizontal θ = 90°
PE = 110 x 2 (1 - cos 90°)
PE = 220 J
now, calculating potential energy when string made 25° with horizontal
PE = m g L (1 - cos θ)
when rope is horizontal θ = 25°
PE = 110 x 2 (1 - cos 25°)
PE = 20.61 J
Answer:
restoring force is 2 N
Explanation:
given data
angle pulled = 5°
force = 1 N
pulled = 10°
to find out
restoring force
solution
we know here force
force = m×g×sinθ ..........1
so here θ is very small so sinθ = θ
1 = mg(5)
mg = 1 /5 ..................2
and
now for 10 degree
we know here θ is small so sinθ = θ
so from equation 1
force = m×g×sinθ
put equation 2 here
force = 1/5 × 10
force = 2 N
so restoring force is 2 N
