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

Answer:

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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Answer 2

The velocity of The second body after collision is 1.65 ms

Given :

The mass of the first body A = m1 = 100g.

The mass of the second body B = m2 = 200g.

Initial velocity of the first body = u1 = 2m s.

Initial velocity of the second body = u2 = 1m s.

Velocity of the first body after collision = v1 = 1.67m s.

To Find ;

Velocity of the second body after collision = v2 = ?

Formula Applied :

$m1v1 + m2 v2 = m1u1 + m2u2$

Solution :

As by the formula,

$m1v1 + m2 v2 = m1u1 + m2u2$

Let's make the grams as kilograms

$mass\:of\: body\: A=\frac{100g}{1000} =0.1kg\\mass\: of\: the\: body\: B = {200}{1000} = 0.2kg$

$0.1 \times 1.67 + 0.2 \times v2 = 0.1 \times 2 + 0.2 \times 1$

$1.67 + 0.2v2 =0.2 +0.2$

$1.67 + 0.2v2 = 0.4$

$0.2v2 = 0.4 - 1.67$

$0.2v2 = 0.23$

$v2 = \frac{0.23}{0.2}$

$v2 = 1.65 ms$

The velocity of the second body after collision is 0.165m s.