Question

In the arrangement show, the pendulum on the left is pulled aside. It is then released and allowed to collide with other pendulum which is at rest. A perfectly inelastic colli-sion occurs and the system rises to a height 1/4 h. The ratio of the masses of the pendulum is :
A= 1
B= 2
C=3
D=4​

Answer 1

Answer:

The answer will be 1

Explanation:

According to the problem the there is a mass A which is located at h height from the ground and got pushed , the result is it collide with the mass B

Therefore by energy conservation theory before collision,

mgh= 1/2mv^2

=> v = √2gh [ velcity of A while colliding]

Therefore during collision the momentum conservation will be,

m(a)v(a) = m(a)+m(b) x v

=> v = m(a) v(a)/m(a) +m(b)

Now the system goes upto h/4

therefore by applying energy conservation,

(m(a)+m(b))g h/4 = 1/2 m(a) m(b) ( m(a) v(a)/m(a) +m(b))^2

now by calculating this we can get , m(a)/m(b) = 1

Answer 2

this is the answer

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