Question

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

ω = \frac{π}{3} rad / s

. Simultaneously at

t = 0

, a small pebble is projected with speed

ν

from point

P

at an angle of

4 5^{\circ}

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}

)

(A) $50 m / s$ ✓ Correct
(B) $51 m / s$
(C) $52 m / s$
(D) $53 m / s$

Answer: (A)

Solution

For the pebble to hit the block at

t = 1

s, the time of flight must equal 1 s. For projectile motion at

4 5^{\circ}

:

\frac{2 v s i n 4 5 ^{\circ}}{g} = 1

. Substituting:

\frac{2 v \cdot \frac{1}{2}}{10} = 1

, giving

\frac{v 2}{10} = 1

, so

v = \frac{10}{2} \cdot 2 = 50

m/s.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option

Accepted Answer

(A) $\sqrt{50} \mathrm{~m} / \mathrm{s}$

Answer

(B) $\sqrt{51} \mathrm{~m} / \mathrm{s}$

Answer

(C) $\sqrt{52} \mathrm{~m} / \mathrm{s}$

Answer

(D) $\sqrt{53} \mathrm{~m} / \mathrm{s}$

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Kinematics

Solution

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