Laws of Motion

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New answer posted•a year ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is __________m.

[Use g = 10m/s²]

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Let ‘h’ be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

$h = u t + \frac{1}{2} a t^{2}$

$= 0 \times 1 + \frac{1}{2} \times 10 \times 1 \times 1$

= 5m

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New answer posted•a year ago

Laws of Motion physics ncert exemplar solutions class 12th chapter two

In the arrangement shown in figure a₁, a₂, a₃ and a₄ are the accelerations of masses m₁, m₂, m₃ and m₄ respectively. Which of the following relation is true for this arrangement?

25 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\sum_{}^{} \vec{T} . \vec{a} = 0$

$\frac{T}{2} a_{1} + \frac{T}{4} a_{2} + \frac{T}{8} a_{3} + \frac{T}{8} a_{4} = 0$

$\Rightarrow 4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0$

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New answer posted•a year ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads^-1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the will then rotate with an angular velocity (in rads^-1).

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

By conservation of Angular momentum

L_i = L_f

$M R^{2} ω = (m R^{2} + 2 m R^{2}) ω'$

$\frac{2 M}{M + 2 m} = ω'$

$\Rightarrow$ $ω' = \frac{2 M}{M + 2 m}$

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New answer posted•a year ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

A block of metal weighing 2kg is resting on a frictionless plane (as shown in figure). It is struck by a jet relasing water at a rate of 1 kgs^-1 and at a speed of 10 ms^-1. Then, the initial acceleration of the block, in ms^-2, will be:

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$a = \frac{F}{m} = \frac{m v}{m} = \frac{1 * 10}{2} = 5 m / s e c^{2}$

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New answer posted•a year ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A body of mass M at rest explodes into pieces, in the ratio of masses 1 : 1: 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms^-¹ and 40 ms-¹ respectively. The velocity of the third piece will be:

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Let ‘v’ be the velocity of third piece

com along x-axis

Upvote

New answer posted•a year ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A body of mass M at rest explodes into pieces, in the ratio of masses 1 : 1: 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms^-1 and 40 ms^-1 respectively. The velocity of the third piece will be

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Let ‘v’ be the velocity of third piece

com along x-axis

M × 0 = M₁v_x + M₃v’_x

$\Rightarrow 0 = M_{1} \times 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} \times 30$

= $\frac{- 1}{2} \times 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

M × 0 = M₂vy + M₃v’y

v_y = $\frac{- M_{2}}{M_{3}} v_{y}$

v_y = $\frac{- 1}{2} \times 40$

v_y = 20m/sec

$\vec{v} = - 15 \hat{i} - 20 \hat{j}$

v = $\sqrt[]{1 5^{2} + 2 0^{2}} = \sqrt[]{225 + 400} = 25 m / s e c$

Upvote

New question posted•a year ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A body of mass M at rest explodes into pieces, in the ratio of masses 1 : 1: 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms-1 and 40 ms-1 respectively. The velocity of the third piece will be

1 Views|Posted a year ago

Asked by Payal Gupta

New answer posted•a year ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

Two masses M₁ and M₂are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M₂is twice that of M₁. The acceleration of the system is a₁. When the mass M₂ is thrice that of M₁ the acceleration of the system is a₂. The ratio $\frac{a_{1}}{a_{2}}$ will be:

15 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$a_{1} = \frac{m_{2} g - m_{1} g}{m_{1} + m_{2}} = \frac{2 m g - m g}{3 m} = \frac{g}{3}$

$a_{2} = \frac{3 m g - m g}{4 m} = \frac{g}{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{2}{3}$

Upvote

New answer posted•a year ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A monkey of mass 50kg climbs on a rope which can withstand the tension (T) of 350 N. If monkey initially climbs down with an acceleration of 4 m/s² and then climbs up with an acceleration of 5 m/s². Choose the correct option (g = 10 m/s²).

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

For climbing downward

50 g – T = 50a

T = 500 – 50 * 4

= 300 N < 350 N

for climbing upwards

T – 50 g = 50 a

T – 500 = 50 * 5

T = 750 N > 350 N

Upvote

New answer posted•a year ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

Three masses M = 100kg, m₁ = 10kg and M₂ = 20kg are arranged in a system as shown in figure. All the surface are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass m₂ moves upward with an acceleration of 2ms^-2. The value of F is:

(Take g = 10ms^-2)

24 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Given Acceleration of m₂ is 2m/s²

(upward)

N = 20a T – m₂g = m₂a

T – 20g = 20a

T = 200 + 20 × 2

T = 240 N

T = m₁ a₁

240 = 10a₁

a₁ = 24 m/s²

$\Rightarrow T a + T a_{1} - T a_{2} = 0$

$a_{2} = 24 + 2 = 26 m / s^{2}$

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