A small block is connected to one end of a massless spring of unstretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t=0. lt then executes simple harmonic motion with angular frequency $ω = \frac{π}{3} r a d / s$ .
Simultaneously at t=0, a small pebble is projected with speed v from point P at an angle of 45 as shown in the figure. Point P is at a horizontal distance of 10 m from O. If the pebble hits the block at t = 1 s, the value of v is
(take $g = 10 m / s^{2}$ )

Question

A small block is connected to one end of a massless spring of unstretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t=0. lt then executes simple harmonic motion with angular frequency

ω = \frac{π}{3} r a d / s

.
Simultaneously at t=0, a small pebble is projected with speed v from point P at an angle of 45 as shown in the figure. Point P is at a horizontal distance of 10 m from O. If the pebble hits the block at t = 1 s, the value of v is
(take

g = 10 m / s^{2}

)

Toppr Admin · Accepted Answer

The block starts its oscillation from its positive extremum-x2234-xblock-t-4-9-0-2cos-x3C9-t-x2234-xblock-t-1s-4-9-0-2cos-x3C0-3-xD7-1-5 m-x2234- Range of pebble -5mSo- v2sin2-x3B8-g-5-x27F9-v-x221A-50 m-s

The block starts its oscillation from its positive extremum.∴xblock(t)=4.9+0.2cosωt∴xblock(t=1s)=4.9+0.2cos(π3×1)=5 m∴ Range of pebble =5mSo, v2sin2θg=5⟹v=√50 m/s

The block starts its oscillation from its positive extremum.
$∴ x_{b l o c k} (t) = 4.9 + 0.2 cos ω t$
$∴ x_{b l o c k} (t = 1 s) = 4.9 + 0.2 cos (\frac{π}{3} \times 1) = 5$ $m$
$∴$ Range of pebble $= 5 m$
So, $\frac{v^{2} sin 2 θ}{g} = 5$
$⟹ v = \sqrt{50}$ $m / s$