Newton Laws of Motion Numerical on Force Class-11 Nootan ISC Physics Solutions

Newton Laws of Motion Numerical on Force Class-11 Nootan ISC Physics Solutions Ch-6. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Newton Laws of Motion Numerical on Force Class-11 Nootan ISC Physics Solutions

Newton Laws of Motion Numerical on Force

Class-11 Nootan ISC Physics Ch-6 Solutions of Kumar and Mittal Nageen Prakashan

Board ISC
Class 11
Subject Physics
Writer Kumar and Mittal
Publication  Nageen Prakashan
Chapter-6 Newton Laws of Motion
Topics Numericals Based on Force
Academic Session 2024-2025

Numericals Based on Force

Ch-6 Newton Laws of Motion Class-11 Nootan ISC Physics Solutions Kumar and Mittal Physics of Nageen Prakashan

Question-1: A ship of mass 3 x 10^7 kg and initially at rest, can be pulled through a distance of 3 m by means of a force of 5 x 10^4 N. The water-resistance is negligible. Find the speed attain by the ship.

Answer- Mass of the ship m = 3 × 10^7 kg , Force on the ship F = 5 x 10^4 N

Using Newton’s second law, F = ma

=> a = 5 x 10^4 / 3 x 10^7

=> 5/3 x 10^-3 m/s²

From 3rd equation of motion, v² – u² = 2as and u = 0

=> v²= 2 x 5x 10^-3 × 3

So => v² = 10^-2

=> v = 10^-1 => 0.1 m/s

Question-2: Two balls A and B of masses 100 g and 250 g connected to the ends of a weightless stretched spring are placed on a smooth horizontal surface. On being released, the ball goes towards west with an initial acceleration of 10 cm/s^2. With what initial acceleration and in which direction will the ball A go?

Answer- For block B we can say, Fb = mb x a

given that, mb = 250 g  = 0.250 kg    a = 10 m/s²

now by above formula

F = ma = 0.250 x 10 = 2.50 N

now we know that both blocks are connected by same spring

So spring force will be same on both blocks

so force block “A” the spring force will be same

F = ma

=> 2.5 = ma x a

given that

ma = 100g = 0.100 kg

now by above equation

2.5 = 0.100 x a

solving above equation

a = 25m/s²

=> so acceleration of block “a” is 25 m/s² towards EAST

Question-3: A force produces an acceleration of 16m/s^2 in a body of mass 0.5 kg and an acceleration of 4.0 m/s^2 in another body. If both the bodies are fastened together then how much acceleration will be produced by this force?

Answer- a1 = 16 m/s²

m1 = 0.5 kg

force = F = m1a1 = 16 x 0.5 = 8N

same force is applied to the second mass; m2

F = m2a2 = 8N

or m2 = 8/a2

m2 = 8/4 = 2kg

total mass = m1 + m2 = 0.5 + 2 = 2.5 kg

force is same as before

So, a = F/M => 8/2.5 => 3.2 m/s²

=> So an acceleration of 3.2 m/s² will be produced b this force.

Question-4: On a smooth table are placed two blocks in contact (i) A horizontal force of 5.0 N is applied on the 20 kg block as shown. State by what force this block presses the 10 kg block. (ii) If the above force is applied on the other side of the 10 kg block then by 5.0 N what force the 20 kg block will press the 10 kg block?

On a smooth table are placed two blocks in contact (i) A horizontal force of 5.0 N is applied on the 20 kg block as shown

Answer- Total mass = 20 + 10 = 30 kg

Total force = 5 N

F =ma

a = 5/30 = 1/6 m/s²

(i) F = ma => (10 x 1/6) = 10/6 N.

(ii) F = ma => (20 x 1/6) = 20/6 N.

—:  end of Newton Laws of Motion Numerical on Force ISC Class-11 Nootan Ch-6 Solutions Kumar and Mittal Physics of Nageen Prakashan :—

Return to : –  Nootan Solutions for ISC Physics Class-11 Nageen Prakashan

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