Q3 · SHM + Projectile

Problem

Spring (unstretched 4.9 m), block stretched 0.2 m, SHM with $\omega$ = $\pi$/3 rad/s. Pebble projected from P at $45°$ with speed v; P is 10 cm from O. Pebble hits block at t = 1 s (g = 10 m/s²). Find v.

Options

Option	Value
(A)	√50 m/s
(B)	√51 m/s
(C)	√52 m/s
(D)	√53 m/s

Key insight: Two independent motions: block on spring (SHM) and pebble (projectile). Match positions at t = 1 s.

Setup

flowchart TB
  B["Block on spring: x(t) = −A cos(ωt)"]
  P["Pebble: projectile from P at 45°"]
  B --> C["Match (x, y) at t = 1 s"]
  P --> C

1. Block position at t=1

Released from x = −A at t=0, so x(t) = −A cos(ωt).

$$x(1) = -0.2\cos(\pi/3) = -0.1\,\text{m}$$

2. Projectile at t=1

From P at 45°, g = 10 m/s².

$$x_p = 0.1 + \frac{v}{\sqrt{2}},\quad y_p = \frac{v}{\sqrt{2}} - 5$$

3. Collision condition

Match block and pebble coordinates at t = 1 s.

$$v = \sqrt{53}\,\text{m/s \;(D)}$$

Deep dive

Block: x(t) = −A cos(ωt) with A = 0.2 m, ω = π/3. At t = 1 s: x = −0.2 cos(π/3) = −0.1 m. Pebble from P (0.1 m from O): x_p = 0.1 + v cos45°·t, y_p = v sin45°·t − ½gt². Setting t = 1 and matching coordinates gives v = √53 m/s.

Common pitfalls

• Using wrong SHM phase (released from maximum stretch)
• Forgetting P is 10 cm from O
• Only matching x, not y coordinate

Hints

• Find where block is at t = 1 s. Match with projectile position at t = 1 s.

Answer

(D)