Question

Calculate the angle 'theta' for which if the particle A is
released, it undergoes perfectly elastic collision
with another identical mass B placed at rest and B
is found to just reach the horizontal position of its
string
21
0
B
(2) 60
)
(1) 30°
(3) 37
(4) 53​

Answer 1

Explanation:

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Answer 2

Answer: Answer is 60°.

Given: Two particle A and B.

To Find: Horizontal position of string.

Step-by-step explanation:

Step 1: The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through which the force acts.

Let the velocity of ball A before hitting ball B is v.

From work energy theorem,

$W_T=\Delta T\\\\mg(2l)(1-cos\theta)=\frac{1}{2} mv^2-0\\\\v^2=4gl(1-cos\theta)$

Step 2: Here is an elastic collision occur between A and B. So after collision B get the velocity v.

Now ball B will start moving up to a certain height. From work energy theorem

$-mgl=0-\frac{1}{2}mv^2\\\\mgl=\frac{1}{2}mv^2\\\\ gl=\frac{1}{2}(4gl)(1-cos\theta)\\\\cos\theta=\frac{1}{2}\\\\\theta = 60$

Hence, correct answer is 60°.

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