A body of mass 1 kg falls freely from lens of 100 m on a platform of mass 3 kg which is mounted on a spring having spring constant K=1.25×106 N/m. The body sticks to the platform and the spring’s maximum compression is found to be x. Given that g=10ms−2, the value of x will be close to
(a)
4 cm
(b)
8 cm
(c)
80 cm
(d)
40 cm
A body of mass 1 kg falls freely from lens of 100 m on a platform of mass 3 kg which is mounted on a spring having spring constant K=1.25×106 N/m. The body sticks to the platform and the spring’s maximum compression is found to be x. Given that g=10ms−2, the value of x will be close to
(a)
4 cm
(b)
8 cm
(c)
80 cm
(d)
40 cm
Question ID - 50316 | SaraNextGen Top Answer
A body of mass 1 kg falls freely from lens of 100 m on a platform of mass 3 kg which is mounted on a spring having spring constant K=1.25×106 N/m. The body sticks to the platform and the spring’s maximum compression is found to be x. Given that g=10ms−2, the value of x will be close to
(a)
4 cm
(b)
8 cm
(c)
80 cm
(d)
40 cm
|
A body of mass 1 kg falls freely from lens of 100 m on a platform of mass 3 kg which is mounted on a spring having spring constant K=1.25×106 N/m. The body sticks to the platform and the spring’s maximum compression is found to be x. Given that g=10ms−2, the value of x will be close to |
|||
|
(a) |
4 cm |
(b) |
8 cm |
|
(c) |
80 cm |
(d) |
40 cm |
Velocity of 1 kg block just before it collides with 3kg block =
=
m/s
Applying momentum conversation just before and just after collision.
1×
=4v⇒v=
m/s
Initial compression of spring
1.25×106x0=30⇒x0≈0
applying work energy theorem,
wg + wsp=KE
⇒40×x+
×1.25×106(02−x2)
=0−
×4×v2
Solving x≈ 4 cm