Question

4. Two objects of masses 100 g and
200 g are moving along the same
line and direction with velocities
of 2ms and 1 ms-1, respectively.​

Answer 1

Step-by-step explanation:

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Answer 2

$\huge\fbox{\dag Answer \dag}$

Before Collision After collision

A B → A B

100 g 200g → 100g 200g

2m/s 1m/s v=?

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

Initial momentum of A=$\frac{100}{1000}$×2=$0.2kgms^-1$

Initial momentum of B=$\frac{200}{1000}$×1=$0.2kgms^-1$

∴ Total momentum of A and B before collision

=0.2+0.2=$0.4kgms^-1$

Let the velocity of A after collision =v

∴ Momentum of A after collision =$\frac{100}{1000}$×v=0.1v

Also, momentum of B after collision=$\frac{200}{1000}$×1.67=$0.334kgms^-1$

∴ 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= $\frac{0.066}{0.1}$=$0.66ms^-1$

$\huge\text{hope \:it\:helps}$