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A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be

$m g \cos θ$

$m g \sin θ$

$m g / \cos θ$

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Explanation

When the whole system is accelerated towards left then pseudo force (ma) works on a block towards right.

For the condition of equilibrium

$m g \sin θ = m a \cos θ$ ⇒ $a = \frac{g \sin θ}{\cos θ}$

∴ Force exerted by the wedge on the block

$R = m g \cos θ + m a \sin θ$

R $= m g \cos θ + m (\frac{g \sin θ}{\cos θ}) \sin θ = \frac{m g (\cos^{2} θ + \sin^{2} θ)}{\cos θ}$

R $= \frac{m g}{\cos θ}$

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An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e. 120 km/h, the stopping distance will be

20 m

40 m

60 m

80 m

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Explanation

The stopping distance, $S \propto u^{2}$ $(∵ v^{2} = u^{2} - 2 a s)$

⇒ $\frac{S_{2}}{S_{1}} = {(\frac{u_{2}}{u_{1}})}^{2} = {(\frac{120}{60})}^{2} = 4$

⇒ $S_{2} = 4 \times S_{1} = 4 \times 20 = 80 m$

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The linear momentum p of a body moving in one dimension varies with time according to the equation p = a + bt² where a and b are positive constants. The net force acting on the body is

A constant

Proportional to t²

Inversely proportional to t

Proportional to t

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Explanation

$F = \frac{d p}{d t} \equiv \frac{d}{d t} (a + b t^{2}) = 2 b t$

∴ $F \propto t$

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A man of weight 80 kg is standing in an elevator which is moving with an acceleration of 6 m/s² in upward direction. The apparent weight of the man will be (g = 10 m/s²)

1480 N

1280 N

1380 N

None of these

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Explanation

The apparent weight of man,

$R = m (g + a) = 80 (10 + 6) = 1280 N$

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N bullets each of mass m kg are fired with a velocity v ms^–1 at the rate of n bullets per second upon a wall. The reaction offered by the wall to the bullets is given by

nmv

$\frac{N m v}{n}$

$n \frac{N m}{v}$

$n \frac{N v}{m}$

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Explanation

Total mass of bullets = Nm, time $t = \frac{N}{n}$

Momentum of the bullets striking the wall = Nmv

Rate of change of momentum (Force) = $\frac{N m v}{t}$ = nmv.

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With what minimum acceleration can a fireman slides down a rope while breaking strength of the rope is $\frac{2}{3}$ of his weight

$\frac{2}{3} g$

$\frac{1}{3} g$

Zero

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Explanation

If man slides down with some acceleration then its apparent weight decreases. For critical condition rope can bear only 2/3 of his weight. If a is the minimum acceleration then,

Tension in the rope $= m (g - a)$ = Breaking strength

⇒ $m (g - a) = \frac{2}{3} m g$ ⇒ $a = g - \frac{2 g}{3} = \frac{g}{3}$

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A ball of mass m moves with speed v and it strikes normally with a wall and reflected back normally, if its time of contact with wall is t then find force exerted by ball on wall

$\frac{2 m v}{t}$

$\frac{m v}{t}$

mvt

$\frac{m v}{2 t}$

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Explanation

For exerted by ball on wall

= rate of change in momentum of ball

= $\frac{m v - (- m v)}{t} = \frac{2 m v}{t}$

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A body of mass 5 kg starts from the origin with an initial velocity $\vec{u} = 30 \hat{i} + 40 \hat{j} m s^{- 1}$ . If a constant force $\vec{F} = - (\hat{i} + 5 \hat{j}) N$ acts on the body, the time in which the y–component of the velocity becomes zero is

5 seconds

20 seconds

40 seconds

80 seconds

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Explanation

$u_{y} = 40 m / s$ , $F_{y} = - 5 N$ , $m = 5 k g$ .

So $a_{y} = \frac{F_{y}}{m} = - 1 m / s^{2}$ (As v = u + at)

∴ $v_{y} = 40 - 1 \times t = 0$ ⇒ $t = 40 \sec$ .

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A ball of mass 0.5 kg moving with a velocity of 2 m/sec strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is

2000 N

1000 N

5000 N

125 N

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Explanation

$F_{a v} = \frac{Δ p}{Δ t} = \frac{m v - (- m v)}{Δ t} = \frac{2 m v}{Δ t} = \frac{2 \times 0 .5 \times 2}{10^{- 3}}$ = 2000 N

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A particle moves in the xy-plane under the action of a force F such that the components of its linear momentum p at any time t are $p_{x} = 2 \cos t$ , $p_{y} = 2 \sin t$ . The angle between F and p at time t is

90°

0°

180°

30°

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Explanation

Given that $\vec{p} = p_{x} \hat{i} + p_{y} \hat{j} = 2 \cos t \hat{i} + 2 \sin t \hat{j}$

∴ $\vec{F} = \frac{d \vec{p}}{d t} = - 2 \sin t \hat{i} + 2 \cos t \hat{j}$

Now, $\vec{F} . \vec{p} = 0$ i.e. angle between $\vec{F} and \vec{p}$ is 90°.

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