JEE Challenger
More from Oscillations
PhyMedAdv 2012 • Paper 1

Collision of Projected Pebble with Oscillating Block on Spring

Consider a small block connected to one end of a massless spring having an un-stretched length of 4.9 m4.9\text{ m}. The opposite end of the spring is held fixed at point OO (see figure). The assembly rests upon a frictionless horizontal surface. The block is pulled by 0.2 m0.2\text{ m} and released from rest at time t=0t = 0, executing simple harmonic motion with an angular frequency ω=π3 rad/s\omega = \frac{\pi}{3}\text{ rad/s}. At the same instant t=0t = 0, a small pebble is launched with speed vv from a point PP at an elevation angle of 45∘45^\circ as depicted. Point PP is located at a horizontal distance of 10 m10\text{ m} from OO. If the pebble collides with the block at t=1 st = 1\text{ s}, determine the value of vv (taking g=10 m/s2g = 10\text{ m/s}^2).

Question Diagram 1

Options

A

50 m/s\sqrt{50}\text{ m/s}

Correct
B

51 m/s\sqrt{51}\text{ m/s}

C

52 m/s\sqrt{52}\text{ m/s}

D

53 m/s\sqrt{53}\text{ m/s}

Topics & Concepts

OscillationsSHMProjectile Motion

Step-by-Step Solution

To find the initial speed vv of the pebble, we analyze the motion of both the block executing simple harmonic motion (SHM) and the pebble executing projectile motion.

1. Motion of the Block

The fixed end of the spring is at the origin O(0,0)O (0, 0). The unstretched length of the spring is l0=4.9 ml_0 = 4.9\text{ m}.

At t=0t = 0, the block is stretched by A=0.2 mA = 0.2\text{ m} from its mean position and released from rest. Therefore, the block undergoes Simple Harmonic Motion along the xx-axis about its mean position xmean=4.9 mx_{\text{mean}} = 4.9\text{ m} with amplitude A=0.2 mA = 0.2\text{ m} and angular frequency ω=π3 rad/s\omega = \frac{\pi}{3}\text{ rad/s}.

The position of the block as a function of time tt is given by: xb(t)=l0+Acos⁡(ωt)x_b(t) = l_0 + A \cos(\omega t)

Substituting the given values: xb(t)=4.9+0.2cos⁡(π3t)x_b(t) = 4.9 + 0.2 \cos\left(\frac{\pi}{3} t\right)

At t=1 st = 1\text{ s}, the horizontal position of the block is: xb(1)=4.9+0.2cos⁡(π3)=4.9+0.2×12=4.9+0.1=5.0 mx_b(1) = 4.9 + 0.2 \cos\left(\frac{\pi}{3}\right) = 4.9 + 0.2 \times \frac{1}{2} = 4.9 + 0.1 = 5.0\text{ m}

2. Motion of the Pebble

The pebble is projected from point P(10,0)P(10, 0) at an angle of 45∘45^\circ with the horizontal towards the origin OO.

The components of initial velocity are:

  • Horizontal velocity: vx=−vcos⁡45∘=−v2v_x = -v \cos 45^\circ = -\frac{v}{\sqrt{2}}
  • Vertical velocity: vz=vsin⁡45∘=v2v_z = v \sin 45^\circ = \frac{v}{\sqrt{2}}

The coordinates of the pebble at time tt are: xp(t)=10−(vcos⁡45∘)t=10−v2tx_p(t) = 10 - \left(v \cos 45^\circ\right) t = 10 - \frac{v}{\sqrt{2}} t zp(t)=(vsin⁡45∘)t−12gt2=v2t−12gt2z_p(t) = \left(v \sin 45^\circ\right) t - \frac{1}{2} g t^2 = \frac{v}{\sqrt{2}} t - \frac{1}{2} g t^2

3. Collision Condition

For the pebble to hit the block at t=1 st = 1\text{ s}, its vertical displacement must bring it to the horizontal ground (z=0z = 0), and its horizontal position must match that of the block:

Vertical Condition:

zp(1)=0z_p(1) = 0 v2(1)−12(10)(1)2=0\frac{v}{\sqrt{2}} (1) - \frac{1}{2} (10) (1)^2 = 0 v2=5  ⟹  v=52 m/s=50 m/s\frac{v}{\sqrt{2}} = 5 \implies v = 5\sqrt{2}\text{ m/s} = \sqrt{50}\text{ m/s}

Horizontal Condition Verification:

Substituting v=52 m/sv = 5\sqrt{2}\text{ m/s} into the expression for xp(1)x_p(1): xp(1)=10−522(1)=10−5=5.0 mx_p(1) = 10 - \frac{5\sqrt{2}}{\sqrt{2}} (1) = 10 - 5 = 5.0\text{ m}

Since xp(1)=xb(1)=5.0 mx_p(1) = x_b(1) = 5.0\text{ m}, the pebble and the block are at the exact same location at t=1 st = 1\text{ s}.

Thus, the initial speed of projection is v=50 m/sv = \sqrt{50}\text{ m/s}.

Correct Option: (A)

PreviousNext