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Class 10 Class 12

Laws of Motion

A car moving with a velocity of 20 ms^-1 stopped at a distance of 40 m. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is

320 m
1280 m
160 m
640 m

160 m

Velocity of car (u) = 20 ms^-1

Distance (s) = 40 m

From Newton's third equation,

$v^{2} - u^{2} = 2 as v^{2} - u^{2} = - 2 as {(0)}^{2} - {(20)}^{2} = - 2 (a) \times 40 400 = 2 \times 40 \times a a = \frac{400}{2 \times 40} a = 5 {ms}^{- 2}$

In the second condition, the velocity becomes twice i.e, u' = 2u

Again from Newton's third equation, we get

${(0)}^{2} - {(2 u)}^{2} = 2 \times (5) \times s s = \frac{{(2 u)}^{2}}{2 \times 5} s = \frac{4 u^{2}}{2 \times 5} s = \frac{4 \times 20 \times 20}{2 \times 5} s = 160 m$

A particle of mass m is driven by a machine that delivers a constant power K watts. If the particle starts from rest, the force on the particle at time t is

$square root of mk over 2 end root t to the power of negative 1 divided by 2 end exponent$
$square root of mk space t to the power of negative 1 divided by 2 end exponent$
$square root of 2 mk end root space t to the power of negative 1 divided by 2 end exponent$
$square root of 2 mk end root space t to the power of negative 1 divided by 2 end exponent$

square root of mk over 2 end root t to the power of negative 1 divided by 2 end exponent

As the machine delivers a constant power
So F, v =constant = k (watts)

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The resultant of two forces, one double the other in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is

120°
60°
90°
150°

120°

F₁ = (F) one force

F₂ = (2F) second force

Let θ is angle between F₁ and F₂

Here angle made by resultant F_R is α with F₁

$∴ α = 90 \tan α = \frac{F_{2} \sin θ}{F_{1} + F_{2} \cos θ} = α \frac{F_{2} \sin θ}{F_{1} + F_{2} \cos θ} = \frac{1}{0} F_{1} + F_{2} \cos θ = 0 2 F \cos θ = - F \cos θ = - \frac{1}{2} θ = 120$

Three blocks A, B, and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a 14 N is applied to the 4 kg block, then the contact force between A and B is

6 N

Given, m_A = 4 kg
m_B = 2 kg
=> m_C =1 kg

So total mass (M) = 4+2+1 = 7 kg
Now, F = Ma
14 = 7a
a=2 m/s²

F-F' = 4a
F' = 14-4x2
F' = 6N

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A car is negotiating a curved road of radius R. The road is banked at angle θ. The coefficient of friction between the tyres of the car and the road is μ_s . The maximum safe velocity on this road is,

A car is negotiating a curved road of radius R. The road is banked at angle and the coefficient of friction between the tyres of car and the road is .
The given situation is illustrated as:

In the case of vertical equilibrium,

N cos = mg + f₁ sin mg = N cos ... (i)
In the case of horizontal equilibrium,

... (ii)
Dividing Eqns. (i) and (ii), we get