Question

A ball of weight 200g accelerates at 40m/s from rest along a straight line path , Distance travelled by ball is 2meter in 20 seconds. Find the force applied and final velocity of ball .
(Drive the formula )
$to \: \: drive : \: \dfrac{m(v - u)}{t}$

2) Ram and Sham have 2 balls , Ram has a ball of 100g and Sham has a ball of 200g .Both throw ball along a straight line, Initial Velocity of ball of 100g was noticed 50m/s after 2 min , and Initial Velocity of ball of 200g was noticed 40m/s after 3min .If final velocity of ball of 100g was 20m/s then find the final velocity of ball of 200g.
(nothing to do with time , drive formula feom newtons 3rd law)
$: \:( f_1)_2 = - (f_2)_1$
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Answer 1

1)

➪ Let us consider 2 bodies of mases :

$\large \rm \: m_1 \: \: and \: \: m_2$

➪ Let 2 bodies collide for time 't'

$\\ \large \rm \: (f_1)_2 = \small \: force \: \: acting \: \: on \: 1 \: \: due \: \: to \: \: 2 \\ \\ \large \rm \: (f_2)_1= \small \: force \: \: acting \: \: on \: 2\: \: due \: \: to \: \: 1$

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Suppose an objest of mass 'm' is moving along a straight line path eith Initial Velocity as 'u' .

• Let it accelerate uniformly to Velocity 'v'
• take time 't'

$\large \sf \: \\ \large \rm \: p_1 = mu \\ \\ \large \: \rm p_2 = mv$

The change in momentum

$\: \: \: \: \: \: \: \: \: \: \: \: \: \large \: \: \rm\propto \: \: p _2 - p_1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \large \: \: \rm\propto \: \:mv - mu \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \large \: \: \rm\propto \: \:m \times (v - u)$

Rate of change of momentum :

$\\ \large \rm \propto \: \: \frac{m \times (v - u)}{t} \\ \\ \large \rm \: \: \: \: \mapsto applied \: \: force : \\ \\ \sf \: f \propto \: \frac{m(v - u)}{t} \\ \\ \large\underline{ \boxed{ \: f = \: ma}}$

$\large \: as \: \: \rm \: a = \dfrac{v - u}{t}$

Solving :

Force = ma

• Mass = 200g

• Acceleration = 40m/s

$\\ \large \implies \rm \: force = 200 \times 40 \\ \\ \large \therefore \underline{ \underline{ \rm \: force = 8000n}}$

According to Formula -

$\large \: f = ma \\ \\ \implies \: 8000n = m \times \frac{v - u}{t} \\ \\ \implies \: 8000n = 200\times \frac{v - u}{20}$

u = Initial Velocity = 0

• as body startes from rest

$\\ \implies \: 8000n = 200 \times \frac{v - 0}{20} \\ \\ \implies \: 8000n = \cancel{200} \: \: 10 \times \frac{v }{ \cancel{20}} \\ \\ \implies \: 8000n = 10 \times v\\ \\ \implies \: \frac{8000}{10} = v \\ \\ \implies \: v = 800$

Therefore, Final velocity = 800m/s

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$\\ \large \rm \: (f_1)_2 = \small \: force \: \: acting \: \: on \: 1 \: \: due \: \: to \: \: 2 \\ \\ \large \rm \: (f_2)_1= \small \: force \: \: acting \: \: on \: 2\: \: due \: \: to \: \: 1$

• According to Newton's 3rd law :

$: \: \large \rm ( f_1)_2 = - (f_2)_1$

$\leadsto \:( f_1)_2 = \dfrac{m_1(v_1 -u_1 )}{t} \\ \\ \leadsto \:( f_2)_1 = \dfrac{m_2(v_2 -u_2 )}{t} \\ \\ \\ \implies \sf \: \dfrac{m_1(v_1 -u_1 )}{ \cancel t} = \dfrac{m_2(v_2 -u_2 )}{ \cancel t} \\ \\ \implies \sf \: {m_1(v_1 -u_1 )}= m_2(v_2 -u_2 )$

$\\ \: \: \: \: \: \: \: = \: {m_1v_1 -m_1u_1 }= m_2v_2 -m_2u_2 )$

$\\ \large \rm \underline{ \boxed{ \sf\: {m_1v_1 + m_2v _2}= m_1u_1-m_2u_2 )}}$

• Formula Drived

Applying Formula

$ball_1$ = Ball of Ram

$ball_2$ = Ball of Sham

$\implies \rm{m_1v_1 + m_2v _2}= m_1u_1-m_2u_2 \\ \\ \implies 100 \times 50 + 200 \times 40 = 100 \times 20 + 200v_2 \\ \\ \implies \: 5000 + 8000 = 2000 + 200v_2 \\ \\ \implies \: 13000 = 2000 + 200v_2 \\ \\ \implies \: 13000 - 2000 = 200v_2 \\ \\ \implies \: 11000 = 200v_2 \\ \\ \implies \: v_2 = \frac{11000}{200} \\ \\ \implies \: v_2 = \frac{110 \cancel{00}}{2 \cancel{00} } \\ \\ \boxed{ \tt \therefore \: v_2 = 55m {s}^{ - 1} }$

Final Velocity of ball of 200g = 55m/s