15. A ball of mass 200 g strikes on a wall with
speed 10 m s' and rebounds with speed 6 m per second what is the change in momentum of the ball.?
Answers
Answer :
Change in momentum of the body, ∆P = 0.8 kg m/s.
Explanation :
Given :
Mass of the ball, m = 200 g or 0.2 kg.
Velocity of the ball when striking the wall,i.e, the initial velocity of the ball, v = 10 m/s.
Velocity of the ball after rebounding from the wall,i.e, the final velocity of the ball, u = 6 m/s.
To find :
Change in momentum of the body, ∆P = ?
Knowledge required :
We know the formula for momentum (P) of a body, i.e, p = mv.
[Where : p,m and v are the momentum, mass and velocity of the body, respectively].
Now from the formula of momentum, we get that the change in momentum (∆P) of the body, ∆P = Final Momentum (Pᵥ) - Initial Momentum (Pᵤ).
⠀⠀⠀⠀⠀⠀⠀⠀⠀∴ ∆P = Pᵥ - Pᵤ
⠀⠀⠀⠀ ⠀=> ∆P = mv - mu [∴ P = mv]
⠀⠀⠀⠀⠀⠀=> ∆P = m(v - u)
Hence the formula for change in momentum, ∆P = m(v - u).
Solution :
By using the formula for change in momentum of a body, we get :
⠀⠀=> ∆P = m(v - u)
⠀⠀=> ∆P = 0.2(6 - 10)
⠀⠀=> ∆P = 0.2 × (-4)
⠀⠀=> ∆P = -0.8 kg m/s.
∴ Hence the change in momentum of the body is 0.8 kg m/s.
0.8 is your answer.
I hope this is helpful