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Laws of Motion

270239 six forces lying in a plane and forming angles of $60^{\circ}$ relative to one another are applied to the centre of a homogeneous sphere with a mass $m = 6 kg$ . These forces are radially outward and consecutively $1 N, 2 N, 3 N, 4 N, 5 N$ and $6 N$ . The acceleration of the sphere is

1 0

2

1 / 2 m / s^{2}

3

1 m / s^{2}

4

2 m / s^{2}

Explanation:

Find net force and hence, $a = F_{net} / m$

Laws of Motion

270300 A ball of mass $2 kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2 m$ while applying the force and the ball goes upto $2 m$ height further, find the magnitude of the force. $(g = 10 {ms}^{- 2})$

1

20 N

2

22 N

3

4 N

4

16 N

Explanation:

$v^{2} - u^{2} = 2 a s; F - m g = m a$

Laws of Motion

270301 A body of mass $3 kg$ is moving along a straight line with a velocity of $24 {ms}^{- 1}$ . When it is at a point ' $P$ ' a force of $9 N$ acts on the body in a direction opposite to its motion. The time after which it will be at ' $P$ ' again is,

1

8 s

2

16 s

3

12 s

4

24 s

Explanation:

When it reaches the point ' $P$ 'again $S = 0$ ,
Using $a = \frac{F}{m}; S = u t + \frac{1}{2} a t^{2}$

Laws of Motion

270302 A ball of mass $10 gm$ dropped from a height of $5 m$ hits the floor and rebounds to a height of $1.25 m$ . If the ball is in contact with the ground for $0.1 s$ , the force exerted by the ground on the ball is $(g = 10 m / s^{2})$

1

0.5 N

2

1.5 N

3

0.15 N

4

2.5 N

Explanation:

$F = \frac{d p}{d t} = \frac{m (v - t)}{t}$ where $v = \sqrt{2 g h_{2}}, u = \sqrt{2 g h_{1}}$

Laws of Motion

270239 six forces lying in a plane and forming angles of $60^{\circ}$ relative to one another are applied to the centre of a homogeneous sphere with a mass $m = 6 kg$ . These forces are radially outward and consecutively $1 N, 2 N, 3 N, 4 N, 5 N$ and $6 N$ . The acceleration of the sphere is

1 0

2

1 / 2 m / s^{2}

3

1 m / s^{2}

4

2 m / s^{2}

Explanation:

Find net force and hence, $a = F_{net} / m$

Laws of Motion

270300 A ball of mass $2 kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2 m$ while applying the force and the ball goes upto $2 m$ height further, find the magnitude of the force. $(g = 10 {ms}^{- 2})$

1

20 N

2

22 N

3

4 N

4

16 N

Explanation:

$v^{2} - u^{2} = 2 a s; F - m g = m a$

Laws of Motion

270301 A body of mass $3 kg$ is moving along a straight line with a velocity of $24 {ms}^{- 1}$ . When it is at a point ' $P$ ' a force of $9 N$ acts on the body in a direction opposite to its motion. The time after which it will be at ' $P$ ' again is,

1

8 s

2

16 s

3

12 s

4

24 s

Explanation:

When it reaches the point ' $P$ 'again $S = 0$ ,
Using $a = \frac{F}{m}; S = u t + \frac{1}{2} a t^{2}$

Laws of Motion

270302 A ball of mass $10 gm$ dropped from a height of $5 m$ hits the floor and rebounds to a height of $1.25 m$ . If the ball is in contact with the ground for $0.1 s$ , the force exerted by the ground on the ball is $(g = 10 m / s^{2})$

1

0.5 N

2

1.5 N

3

0.15 N

4

2.5 N

Explanation:

$F = \frac{d p}{d t} = \frac{m (v - t)}{t}$ where $v = \sqrt{2 g h_{2}}, u = \sqrt{2 g h_{1}}$

Laws of Motion

270239 six forces lying in a plane and forming angles of $60^{\circ}$ relative to one another are applied to the centre of a homogeneous sphere with a mass $m = 6 kg$ . These forces are radially outward and consecutively $1 N, 2 N, 3 N, 4 N, 5 N$ and $6 N$ . The acceleration of the sphere is

1 0

2

1 / 2 m / s^{2}

3

1 m / s^{2}

4

2 m / s^{2}

Explanation:

Find net force and hence, $a = F_{net} / m$

Laws of Motion

270300 A ball of mass $2 kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2 m$ while applying the force and the ball goes upto $2 m$ height further, find the magnitude of the force. $(g = 10 {ms}^{- 2})$

1

20 N

2

22 N

3

4 N

4

16 N

Explanation:

$v^{2} - u^{2} = 2 a s; F - m g = m a$

Laws of Motion

270301 A body of mass $3 kg$ is moving along a straight line with a velocity of $24 {ms}^{- 1}$ . When it is at a point ' $P$ ' a force of $9 N$ acts on the body in a direction opposite to its motion. The time after which it will be at ' $P$ ' again is,

1

8 s

2

16 s

3

12 s

4

24 s

Explanation:

When it reaches the point ' $P$ 'again $S = 0$ ,
Using $a = \frac{F}{m}; S = u t + \frac{1}{2} a t^{2}$

Laws of Motion

270302 A ball of mass $10 gm$ dropped from a height of $5 m$ hits the floor and rebounds to a height of $1.25 m$ . If the ball is in contact with the ground for $0.1 s$ , the force exerted by the ground on the ball is $(g = 10 m / s^{2})$

1

0.5 N

2

1.5 N

3

0.15 N

4

2.5 N

Explanation:

$F = \frac{d p}{d t} = \frac{m (v - t)}{t}$ where $v = \sqrt{2 g h_{2}}, u = \sqrt{2 g h_{1}}$

Laws of Motion

270239 six forces lying in a plane and forming angles of $60^{\circ}$ relative to one another are applied to the centre of a homogeneous sphere with a mass $m = 6 kg$ . These forces are radially outward and consecutively $1 N, 2 N, 3 N, 4 N, 5 N$ and $6 N$ . The acceleration of the sphere is

1 0

2

1 / 2 m / s^{2}

3

1 m / s^{2}

4

2 m / s^{2}

Explanation:

Find net force and hence, $a = F_{net} / m$

Laws of Motion

270300 A ball of mass $2 kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2 m$ while applying the force and the ball goes upto $2 m$ height further, find the magnitude of the force. $(g = 10 {ms}^{- 2})$

1

20 N

2

22 N

3

4 N

4

16 N

Explanation:

$v^{2} - u^{2} = 2 a s; F - m g = m a$

Laws of Motion

270301 A body of mass $3 kg$ is moving along a straight line with a velocity of $24 {ms}^{- 1}$ . When it is at a point ' $P$ ' a force of $9 N$ acts on the body in a direction opposite to its motion. The time after which it will be at ' $P$ ' again is,

1

8 s

2

16 s

3

12 s

4

24 s

Explanation:

When it reaches the point ' $P$ 'again $S = 0$ ,
Using $a = \frac{F}{m}; S = u t + \frac{1}{2} a t^{2}$

Laws of Motion

270302 A ball of mass $10 gm$ dropped from a height of $5 m$ hits the floor and rebounds to a height of $1.25 m$ . If the ball is in contact with the ground for $0.1 s$ , the force exerted by the ground on the ball is $(g = 10 m / s^{2})$

1

0.5 N

2

1.5 N

3

0.15 N

4

2.5 N

Explanation:

$F = \frac{d p}{d t} = \frac{m (v - t)}{t}$ where $v = \sqrt{2 g h_{2}}, u = \sqrt{2 g h_{1}}$