SATHEE: Laws Of Motion Question 243

Laws Of Motion Question 243

Question: A body of mass 2 kg has an initial velocity of 3 meters per second along OE and it is subjected to a force of 4 N in a direction perpendicular to OE. The distance of the body from O after 4 seconds will be [CPMT 1976]

Options:

A) 12 m

B) 20 m

C) 8 m

D) 48 m

Show Answer

Answer:

Correct Answer: B

Solution:

Displacement of body in 4 sec along OE $s_{x} = v_{x} t = 3 \times 4 = 12 m$

Force along OF (perpendicular to OE) = 4 N $∴$ $a_{y} = \frac{F}{m} = \frac{4}{2} = 2 m / s^{2}$

Displacement of body in 4 sec along OF therefore $s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$

$= \frac{1}{2} \times 2 \times {(4)}^{2} = 16 m$ [As $u_{y} = 0$ ]

$∴$ Net displacement $s = \sqrt{s_{x}^{2} + s_{y}^{2}} = \sqrt{{(12)}^{2} + {(16)}^{2}} = 20 m$

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Laws Of Motion Question 243

Question: A body of mass 2 kg has an initial velocity of 3 meters per second along OE and it is subjected to a force of 4 N in a direction perpendicular to OE. The distance of the body from O after 4 seconds will be [CPMT 1976]

Options:

Answer:

Solution:

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