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

Given : The mass [ m ] of the ball is 200 g , Initial Velocity [ v ] of the ball is 10 m/s [ when , the ball strikes to wall ] & Final Velocity [ u ] of the ball is 6 m/s [ when , the ball strikes back or rebounds ].

Exigency To Find : Change in momentum of the ball [ [tex ] \triangle [/tex] P ] .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Finding Change in momentum :

Given that ,

• Initial Velocity [ v ] of the ball is 10 m/s [ when , the ball strikes to wall ]
• Final Velocity [ u ] of the ball is 6 m/s [ when , the ball strikes back or rebounds ].
• The mass [ m ] of the ball is 200 g

⠀⠀⠀⠀⠀ Converting mass from g to kg :

$\qquad :\implies \sf Mass \: = 200 g \:$

$\qquad :\implies \sf Mass \: [ \ m \ ]\: = 0.2 kg \qquad \bigg\lgroup \sf{ 1 \ kg = 1000 \ g }\bigg\rgroup \:$

$\qquad :\implies \bf Mass \:[ \ m \ ] \ = 0.2 \ g \:$

Now,

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \maltese \: \: \bf Formula\: of \: Momentum \: [\ P \ ]\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ P = mv }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , P is the momentum, m is the mass & v is the velocity .

Therefore,

$\qquad \maltese \: \: \bf Formula\: of \: Change \: in \: Momentum \: [\ \triangle \ P \ ]\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ \triangle P = P_v - P_u }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , $\bf P_v$ is the Final Momentum , $\bf P_u$ is the Initial momentum & $\triangle$ P is the Change in momentum .

$\qquad :\implies \sf \triangle P = P_v - P_u \:$

$\dag\:\:\sf{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{P \:= \: mv }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , P is the momentum, m is the mass & v is the velocity .

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Applying \: this \::}}$

$\qquad :\implies \sf \triangle P = P_v - P_u \:$

$\qquad :\implies \sf \triangle P = mv - mu \:$

$\qquad :\implies \sf \triangle P =m(v - u) \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad :\implies \sf \triangle P =m(v - u) \:$

$\qquad :\implies \sf \triangle P =0.2(6 - 10) \:$

$\qquad :\implies \sf \triangle P =0.2(-4) \:$

$\qquad :\implies \bf \triangle P = - 0.8 \:$

$\qquad:\implies \frak{\underline{\purple{\:\triangle P = - 0.8\: kg \ m / s }} }\:\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Change \:in \:momentum \:of\:the\:ball \:is\:[ \ \triangle P \ ]\bf{ - 0.8\: kg \ m / s }}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀