if a 20kg mass , when hung vertically from end of a spring, stretches the spring down by 12cm, find the spring constant of the s
1 answer:
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Answer:
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 .