Question

two objects of masses 100g and 200g moving along the same line and direction with velocities of 2m 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.

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Answer 1

Mass of one of the objects, m1 = 100 g = 0.1 kg

Mass of the other object, m2 = 200 g = 0.2 kg

Velocity of m1 before collision, v1= 2 m/s

Velocity of m2 before collision, v2= 1 m/s

Velocity of m1 after collision, v3= 1.67 m/s

Velocity of m2 after collision= v4

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

Therefore, m1v1 + m2v2 = m1v3 + m2v4

2(0.1) + 1(0.2) = 1.67(0.1) + v4(0.2)

0.4 = 0.167 + 0.2v4

v4= 1.165 m/s

Hence, the velocity of the second object becomes 1.165 m/s after the collision.

Answer 2

Answer:

ANSWER IS

$0.66m {s}^{ - 1}$

STEP BY STEP EXPLANATION

Before Collision

A . . . . . B

100 g 200g

2m/s 1m/s

1.67ms

−1

After collision

A . . . . B

→ 100g 200g

v=? 1.67ms

Let the 100 g and 200 g objects be A and B as shown in above figure.

Initial momentum of A=

$\sf{\bold{ \frac{100}{1000}}} \times 2 = 0.2kgm {s}^{ - 1}$

Initial momentum of B=

$\sf{\bold{ \frac{200}{1000}}} \times 1 = 0.2kgm {s}^{ - 1}$

∴Total momentum of A and B before collision =0.2+0.2=

$\sf{\bold{0.4kmg {s}^{ - 1} }}$

Let the velocity of A after collision =v

∴ Momentum of A after collision =

$\sf{\bold{ \frac{100}{1000} }} \times v = 0.1v$

Also, momentum of B after collision

$\sf{\bold{ \frac{200}{1000} }} \times 1.67=0.334kgm {s}^{ - 1} </p><p></p><p>$

∴ Total momentum of A and B after collision

=0.1×v+0.334

Using the law of conservation of momentum, momentum of A and B after collision = momentum of A and B before collision

0.1×v+0.334=0.4

0.1×v=0.4−0.334

⇒v=

$\sf{\bold{ \frac{0.066}{0.1} }} = 0.66m {s}^{ - 1}$

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