(a) 0.02 m
(b) 0.5 m
(c) 0.2 m
(d) 0.05 m
Doubt by Jaskirat
Solution :
mass (m) = 1 kg
speed (v) = 2 m/s
Force constant (k) =100 N/m
As per the law of conservation of energy
The K.E. of the ball will be converted in to the P.E. of the Spring.
KE=PE
½mv²=½kx²
(mv²)/k=x²
x=√[(mv²)/k]
x=√[(mv²)/k]
x=√[(1×(2)²]/100
x=√(4/100)
x=2/10
x=0.2 m
x=√[(mv²)/k]
x=√[(mv²)/k]
x=√[(1×(2)²]/100
x=√(4/100)
x=2/10
x=0.2 m
Hence, (c) 0.2 m, would be the correct option.
Similar Question :
A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal surface collides with a nearly weightless spring of force constant k=50 N/m. The maximum compression of the spring would be :
(a) 0.15 m
(b) 0.12 m
(c) 1.5 m
(d) 0.5 m
(a) 0.15 m
(b) 0.12 m
(c) 1.5 m
(d) 0.5 m