Question

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.?​

Answer 1

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.

Answer 2

$\Huge\sf\underline{\underline{\blue{Answer}}}$

Mass of ball(m)=200g=0.2kg

Initial velocity of ball(u)=10ms-¹

Final velocity of ball(v)=5ms-¹

Therefore, change in momentum (△p) = Final momentum - initial momentum =m(v−u)=0.2×(−5−10)=−3 kgms-¹

=−3 Ns