1. An object (m = 500 g) with an initial speed of 0.2 m/s collides with another object (m = 1.5 kg) which was at rest before the collision. Calculate the resulting speed for an inelastic collision (when they stick together). 2. A small object (m = 200 g) collides elastically with a larger object (m = 1000 g), which was at rest before the collision. The incoming speed of the smaller object was 1.0 m/s. The speed of the larger object after the collision is 0.33 m/s. Calculate the resulting speed and determine the direction for the smaller object after the collision when it rebounds. (Watch out for the directions of the motions and use respective signs for the velocities and momentums.)

Explanation:

1 )

We shall apply conservation of momentum law to solve the problem.

mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .

.5 x .2 = (1.5 + .5)V

V = .05 m /s

2 ) We shall use formula for velocity of object after elastic collision as follows

v₁ =

m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.

Putting the values

=

= - .66 m /s

Since the sign is negative so it will be in opposite direction .

step-by-step explanation:

we are given

we have to simplify it

firstly, we will factor it

now, we can cancel like terms

now, we can combine it

so, we get

ice to water 1 water to vapour 2 latent heat