Question

A body of mass 20 kg moving with a speed of 10 ms -1 on a horizontal smooth surface
collides with a massless spring of spring constant 5 N/m. If the mass stops after
collision, distance of compression of the spring will be :
(a)
10 m
(b)
50 m
(c)
5 m
(d)
20 m​

Answer 1

Answer:

D

Explanation:

As the body stops , it's total kinetic energy is converter into spring potential energy

1/2mv^2 = 1/2 kx^2

0.5x20x(10)^2 =0.5 x 5 x (x)^2

400 =x^2

X = 20m

Answer 2

The distance of compression of spring will be (d)20m

1.Loss in K.E = Gain In potential elastic energy

initial K.E=1/2mv²

2.1/2 * 20*10*10=1000

final K.E=0

3.1/2kx²=loss in K.E

1/2kx²=1000

therefore x=20m