Question

A ball of mass 100g and another ball of 150 g moves towards each other with

speeds 10m/s and 5m/s respectively. If they stick to each other after colliding,

what would be the velocity of the combined as after the collision?​

Answer 1

Answer:

• Velocity of the combination as after the collision = 1 m/s

Explanation:

Given

• Mass of first ball, m = 100 g = 0.10 kg
• Initial velocity of the first ball, u = 10 m/s
• Mass of second ball, M = 150 g = 0.15 kg
• Initial velocity of the second ball, U = -5 m/s    [Negative because balls are moving towards each other]
• Balls stick to each other after colliding

To find

• The Velocity of the combined balls as after the collision, V =?

Knowledge required

• Law of conservation of linear momentum

For two bodies of masses m and M moving with initial velocities u and U and after the collision, their final velocities became v and V., then

       m u + M U = m v + M V

Solution

Using the law of conservation of linear momentum

[ Since balls stuck to each other after colliding, therefore final velocities of both balls would be same that is 'V' ]

→ m u + M U = m V + M V

→ m u + M U = V ( m + M )

→ 0.10 × 10 + 0.15 × -5 = V ( 0.10 + 0.15 )

→ 1.00 - 0.75 = 0.25  V

→ 0.25 = 0.25  V

→ V = 0.25 / 0.25

→ V = 1 m/s

Therefore,

• The final velocity of the combination will be 1 m/s in the direction of initial movement of the first ball.

Answer 2

Given -

• v₁ = velocity of ball going right = 10 m/s

• V₂ = velocity of ball going left = - 5 m/s

• m₁ = mass of ball going right = 100 g = 0.1kg

• M₂ = mass of ball going left = 150 g = 0.15kg

To Find :

• what would be the velocity of the combined as after the collision?

Solution :

concept :

Newton's third law: To every action, there is an equal and opposite reaction. The action and reaction acts on different bodies at the same time and don't cancel each other.

_____________________

$: \implies \: \: \: \boxed{ \sf \: (m_{1}) (u_{1}) + \sf \: (m_{2}) (u_{2}) = (m_{1} + m_{2})v}$

Substitute all values :

$: \implies \: \: \: \sf \:(0.1 \times 10) + (0.15 - 5) = (0.1 + 0.15)v \\ \\ \\ :\implies \: \: \: \sf \:1 - 0.75 = 0.25v \\ \\ \\ :\implies \: \: \: \sf \:0.25 = 0.25v \\ \\ \\ :\implies \: \: \: \sf \: \: v = \cancel{\frac{0.25}{0.25} } \\ \\ \\ :\implies \: \: \: \sf \:v = 1$

• 1 m/s is the velocity of the combined as after the collision

More to know :

State the law of conservation of momentum.

• The law of conservation of momentum states that for two objects colliding in an isolated system, the total momentum before and after the collision is equal.