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. It then executes simple harmonic motion with angular frequency $\omega = \pi/3$ rad/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 = 1s, the value of $v$ is? (take, g = 10 m/s²)

• A. $\sqrt{50}$ m/s
• B. $\sqrt{51}$ m/s
• C. $\sqrt{52}$ m/s
• D. $\sqrt{53}$ m/s

Answer 1

Answer: (A)

$\sqrt{50}$ m/s

Answer 2

The block's position is given by $x_{block}(t) = x_{eq} + A \cos(\omega t)$. With $x_{eq} = 4.9$ m, $A = 0.2$ m, and $\omega = \pi/3$ rad/s, at $t=1$ s, $x_{block}(1) = 4.9 + 0.2 \cos(\pi/3) = 4.9 + 0.2(1/2) = 5.0$ m.

The pebble's motion equations are $x_{pebble}(t) = 10 - (v/\sqrt{2})t$ and $z_{pebble}(t) = (v/\sqrt{2})t - (1/2)gt^2$. For the pebble to hit the block at $t=1$ s, $z_{pebble}(1) = 0$. $(v/\sqrt{2})(1) - (1/2)(10)(1)^2 = 0 \implies v/\sqrt{2} = 5$. Thus, $v = 5\sqrt{2} = \sqrt{50}$ m/s. At $t=1$ s, $x_{pebble}(1) = 10 - (5)(1) = 5$ m, which matches $x_{block}(1)$.