SATHEE: Simple Harmonic Motion 4 Question 5

Simple Harmonic Motion 4 Question 5

5. 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 $ω = \frac{π}{3} r a d / s$ . Simultaneously at $t = 0$ , a small pebble is projected with speed $v$ from point $P$ at an angle of $45^{\circ}$ 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, $g = 10 m / s^{2}$ )

(2012)

(a) $\sqrt{50} m / s$

(b) $\sqrt{51} m / s$

(d) $\sqrt{53} m / s$

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Answer:

Correct Answer: 5. (a)

Solution:

$t$ =time of flight of projectile

$\begin{aligned} = \frac{2 v \sin θ}{g} (θ = 45^{\circ}) \\ ∴ v = \frac{g t}{2 \sin θ} & = \frac{10 \times 1}{2 \times 1 / \sqrt{2}} \\ = \sqrt{50} m / s \end{aligned}$

Simple Harmonic Motion 4 Question 5

Answer:

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Simple Harmonic Motion 4 Question 5

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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