Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms–1 and 1 ms–1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms–1. Determine the velocity of the second object.
Answers
Required Answer:
Given:
Before collision:
- Mass of the 1st object (m_1) = 100g = 0.1 kg
- Mass of the 2nd object (m_2) = 200g = 0.2kg
- Velocity of the 1st object before collision ( v_1 ) = 2m/s
- Velocity of the 2nd object before collision ( v_2 ) = 1m/s
After collision:
- Velocity of the 1st object after collision ( v_1 ) = 1.67m/s
To calculate:
- Velocity of the second object after collision ( v_2 )
Calculation:
According to the law of conservation of momentum:
- Total Momentum before collision = Total Momentum after collision
Let us calculate the momentum first:
Before collision:
◆ Momentum of the first object before collision:
→ P = mv
Where,
• P , momentum = ?
• m , mass = 0.1 kg
• v, velocity = 2m/s
|| Substituting values: ||
→ P = 0.1 × 2
→ P = 0.2 kg m/s
Therefore, momentum of the first object before collision is 0.2 kg m/s.
◆ Momentum of the second object before collision:
→ P = mv
→ P = 0.2 × 1
→ P = 0.2 kg m/s
Therefore, momentum of the second object before collision is 0.2 kg m/s.
● Total momentum before collision = momentum of the first object + momentum of the second object
→ Total momentum before collision = ( 0.2 + 0.2 )kg m/s
→ Total momentum before collision = 0.4kg m/s
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After collision:
◆ Momentum of the first object after collision:
→ P = mv
Where,
• P , momentum = ?
• m , mass = 0.1 kg
• v, velocity = 2m/s
|| Substituting values: ||
→ P = 0.1 × 1.67
→ P = 0.167 kg m/s
Therefore, momentum of the first object before collision is 0.167 kg m/s.
◆ Momentum of the second object after collision:
→ P = mv
→ P = 0.2 × v
→ P = 0.2v ( Let the velocity be v )
Therefore, momentum of the first object before collision is 0.2v kg m/s.
● Total momentum after collision = momentum of the first object + momentum of the second object
→ Total momentum after collision = 0.167 + 0.2v
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Now, according to the law of conservation of momentum:
- Total Momentum before collision = Total Momentum after collision
→ 0.4 = 0.167 + 0.2v
→ 0.2v = 0.4 - 0.167
→ 0.2v = 0.233
→ v = 0.2333/0.2
→ v = 1.165 m/s
Therefore, velocity of the 2nd object after collision is 1.165 m/s.
Let us consider there is no external unbalanced force acting in the given system, as the Law of Conservation of Momentum is not valid in case of any external force.
We have,
Mass of the objects = 100 g (m); 200 g (M).
Intial velocities respectively = 2 m/s (u); 1 m/s (U).
Final velocity of the first object = 1.67 m/s (v).
Let, final velocity of the second object = (V) m/s
Formula of conservation of Momentum:-
mu + MU = mv + MV
{Momentum before collision is same as that of after collision.}
Before collision:-
mu + MU
= (100 g)(2 m/s) + (200 g)(1 m/s)
= 200 g m/s + 200 g m/s
= 400 g m/s
After collision:-
mv + MV
= (100 g)(1.67 m/s) + 200V
= 167 g m/s + 200V
∴ 167 g m/s + 200V = 400 g m/s
⇒ 200V = 400 g m/s - 167 g m/s = 233 g m/s
⇒ V = 1.165 m/s {Answer}