Question

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Answer 1

Question :

Two objects 100g and 200 g are moving along the same line and direction with velocities 2m/s and 1 m/s , respectively . They collide and after collision ,the first object moves at a velocity of 1.67m/s . Determine the velocity of the second object.

Theory :

•Momentum:

the linear momentum of a body is defined by the product of the mass of the body and its velocity.i.e

$\sf \: \vec{p } = m \vec{v }$

•law of conservation of linear momentum

if the total external force acting on a system is equal to zero,then the final value of the momentum of the system is equal to the initial value of the total momentum of the system.

$\sf \: \vec{F _{ext}} = \dfrac{d \vec{p}}{dt}$

$\sf \: if \: \vec{F_{ext}} = 0 \: i.e \: \dfrac{ d \vec{p}}{dt} = 0$

$\sf \: then \: \vec{p} = constant \: or \: \vec{p_{f}} = \vec{p_{i}}$

Solution :

Given :$\sf \: m_{1} = 100g = 0.1kg$

$\sf \: m_{2} = 200g = 0.2kg$

$\sf \: u_{1} = 2ms {}^{ - 1}$

$\sf \: u_ {2} = 1ms {}^{ - 1}$

$\sf \: and \: v_{1} = 1.67ms {}^{ - 1}$

We have to find the value of object 2

By law of conversation of momentum:

Total momentum before collision= Total momentum after the collision

$\implies \sf \: m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$

Now put the given values

$\sf \implies0.1 \times2 + 0.2 \times 1 = 0.1 \times 1.67 + 0.2v_{2}$

$\implies \sf0.4 = 0.2v_{2} + 0.167$

$\implies \sf0.2v_{2} = 0.4 - 0.167$

$\sf \implies 0.2v_{2} = 0.233$

$\sf \implies \: v_{2} = \frac{ \cancel{0.233}}{ \cancel{0.2}} = 1.165ms {}^{ - 1}$

$\sf \implies \: v_{2} = 1.165ms {}^{ - 1}$

Therefore,the velocity of Object 2 after the collision is 1.165 m/s .

Answer 2

Question:-

Two objects of masses 100 grams and 200 grams are moving along the same lind and direction with velocities 2 m/s and 1 m/s respectively. They collide and after collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.

Answer:-

Let us assume that the system is free from any external unbalanced force.

Mass of object A = 100 g $\tt{ (m_a) }$

Initial velocity of A = 2 m/s $\tt{ (u_a) }$

Mass of object B = 200 g $\tt{ (m_b) }$

Initial velocity of B = 1 m/s $\tt{ (u_b) }$

As per the law of conservation of momentum,

$\boxed{\tt{ {m}_{a}{u}_{a} + {m}_{b}{u}_{b} = {m}_{a}{v}_{a} + {m}_{b}{v}_{b}} }$

Or, momentum before collision and after collsion remains same.

LHS:-

(100 g)(2 m/s) + (200 g)(1 m/s)

= 200 g m/s + 200 g m/s

= 400 g m/s

RHS:-

(100 g)(1.67 m/s) + (200 g)($\tt{ v_b }$)

= 167 g m/s + (200 g)($\tt{ v_b }$)

∵ LHS = RHS,

∴ 167 g m/s + (200 g)($\tt{ v_b}$) = 400 g m/s

→ (200 g)($\tt{ v_b }$) = (400 - 167) g m/s

→ $\tt{ v_b }$ = (233 g m/s)/(200 g)

→ $\tt{v_b}$ = 1.165 m/s

∴ The velocity of the second object after collision become 1.165 m/s.