A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30° with the original direction. The ratio of velocities of the balls after collision is x ∶ y, where x is __________.

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JEE Main 04 April 2024 Shift 1
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Detailed Solution

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Explanation: 

In this elastic collision is used. It is a collision when the combined kinetic energy of the two bodies does not change.

Concept: We can understand the situation using a diagram :

 F1 Madhuri UG Entrance 20.10.2022 D17 

where we have explained position of balls before and after collision.

conservation of momentum can be used because momentum remains conserved in both x and y direction so that:

initial momentum = final momentum   (along y axis)

Calculation:

Given: 

initial speed of ball 1 = V1 = 9 m/s

initial speed of ball 2 = V2 = 0 m/s

angle made by diffraction of ball with axis = 30o

Applying conservation of momentum along y-axis:

Pi = initial momentum = 0 and Pf = final momentum = mV1 sin(30) ( î ) + mV2 sin(30) (-î )

Here, Pi = PF ⇒ 0 = mV1 sin(30) (î ) + mV2 sin(30) (-î ) 

mV1 sin(30) (î ) = mV2 sin(30) (î)

∵ V= V2  so the velocity ratio is,  V1 : V2 = 1: 1 

hence, x = 1 and y = 1

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