IIT-JEE 2012 · Paper 1 · Physics Q5

← IIT-JEE 2012 · Paper 1

PhysicsSingle correct+3 / -1Question 5 of 60

A small block is connected to one end of a massless spring of un-stretched 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 $\omega=\dfrac\pi3$ 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 $t=1$ s, the value of $v$ is (take $g=10\ \mathrm{m/s^2}$ )

A spring fixed at the wall O on the left holds a block on a horizontal x-axis; a pebble at P, 10 m from O, is projected with speed v at 45° above the horizontal, back towards the block (z-axis vertical).

\sqrt{50}

m/s

\sqrt{51}

m/s

\sqrt{52}

m/s

\sqrt{53}

m/s

Answer and worked solution

Official answer: A

Block at $t=1$ s. $x=4.9+0.2\cos\left(\frac\pi3\cdot1\right)=4.9+0.1=5.0$ m from O. So the pebble must travel $10-5=5$ m horizontally in 1 s.

Pebble. Time of flight 1 s: $\dfrac{2v\sin45°}g=1$ , so $v\sin45°=5$ m/s. Its horizontal speed $v\cos45°=5$ m/s also gives 5 m in 1 s. Consistent.

Speed. $v=5\sqrt2=\sqrt{50}$ m/s.

(A)

Topics:Spring-mass systems Projectile motion Simple harmonic motion

← Previous question Next question →

IIT-JEE 2012 · Paper 1, Physics question 5