Question

${\huge{\bf{Question}}}$
Two objects of masses 100g and 200g are moving along the same line and direction and velocities of 2m/s -1 and 1m/s -1, respectively .They collide and after the collision, the first object moves at a velocity of 1.67 m/s -1. Determine the velocity of the other object.


Note :
1.No copy paste and no screenshots
2.No spams
3.Only high quality answers.​

Answer 1

Required Solution :-

Given -

•Mass of 1st object object $\sf {(m_1) }$= 100 g

→ after converting into kg

→ 100/1000

→ 0.1 kg

• Mass of 2nd object $\sf {(m_2) }$= 200g

→ after converting into kg

→ 200/1000

→ 0.2 kg

• Velocity of 1st object $\sf {(u_1) }$ =2m/s-¹

• Velocity of 2nd object $\sf {(u_2) }$ = 1m/s-¹

• After Collison velocity of 1st object $\sf {(v_1) }$ = 1.67m/s

To Find -

• After Collison velocity of 2nd object $\sf {(v_2) }$ = ?

Solution -

Now we know that ,

$:\implies\sf{FBA = -FAB }$

$:\implies\sf{ {m}_{2} \times {a}_{2} = - {m}_{1} \times {a}_{1} }$

we know that a = v - u / t

$\sf:\implies{ {m}_{2} \times \dfrac{ {v}_{2} - {u}_{2}} {t} = - {m}_{1} \times\dfrac{ {v}_{1} - {u}_{1}} {t} }$

$\sf:\implies{ \dfrac{ {m}_{2}{v}_{2} -{m}_{2} {u}_{2}} {t} = - \dfrac{-{m}_{1} {v}_{1} + {m}_{1} {u}_{1}} {t} }$

★ t will be cancelled

$\sf:\implies{ { {m}_{2}{v}_{2} -{m}_{2} {u}_{2}} = {-{m}_{1} {v}_{1} + {m}_{1} {u}_{1}} }$

hence it will be

$\sf:\implies{\bold{ {m}_{1} {u}_{1} + {m}_{2} {u}_{2} = {m}_{1} {v}_{1} + {m}_{2}{v}_{2} }}$

♪ law of conservation of momentum,

→ Total momentum before collision = Total momentum after collision

Now by applying the formula !

$\sf:\implies{ 0.1kg \times 2m/s + 0.2kg \times 1m/s = 0.1 kg \times 1.67 m/s + 0.2kg \times {v}_{2} }$

$\sf:\implies{ 0.4 kg \; m/s = 0.167 kg \; m/s + 0.2kg \times {v}_{2} }$

$\sf:\implies{ 0.4 kg \; m/s - 0.167 kg \; m/s = 0.2kg \times {v}_{2} }$

$\sf:\implies{ {v}_{2} = \dfrac{0.233 kgm/s }{0.2 kg} }$

$\sf:\implies{{v}_{2} = \bold{1.165m/s} }$

hence , the velocity of other object will be 1.165 m/s