Question

A bullet of mass 20g strikes a block of 980g with velocity v.the spring constant is 100.after collision the spring is compressed upto 10cm.find velocity of block after collision

Answer 1

0.5mv^2 = 0.5kx^2
v = x * sqrt(k/m)
v = 10 * 10^-2 * sqrt(100 / 1)
v = 1 m/s

Velocity of block after collision is 1 m/s

Answer 2

H e y a ! ! !

YOUR ANSWER - 1 m / s

SOLUTION -:

0.5mv^2 = 0.5kx^2
v = x × sqrt(k/m)

v = 10 × 10^-2 × sqrt(100 / 1)

v = 1 m/s

Hence , the Velocity of block after collision is 1 m/s.. .

#hope it helps !