Question

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Answer 1

Question : -

A ball of mass 10 g is moving with a velocity of 50 m/s . On applying a constant force on ball for 2.0 seconds , it acquires a velocity of 70 m/s . Calculate :

(i) the initial momentum of ball,

(ii) the final momentum of ball,

(iii) the rate of change of momentum,

(iv) the acceleration of ball, and

(v) the magnitude of force applied.

ANSWER

Given : -

A ball of mass 10 g is moving with a velocity of 50 m/s . On applying a constant force on ball for 2.0 seconds , it acquires a velocity of 70 m/s .

Required to find : -

• (i) the initial momentum of ball,
• (ii) the final momentum of ball,
• (iii) the rate of change of momentum,
• (iv) the acceleration of ball, and
• (v) the magnitude of force applied.

Formulae used : -

Momentum (p) = Mass (m) x Velocity (v)

Rate of change of momentum (∆p) = (mv - mu)/t

Force (F) = Mass (m) x Acceleration (a)

Equations used : -

v = u + at

Here,

v = final velocity

u = initial velocity

a = acceleration

t = time taken

Solution : -

A ball of mass 10 g is moving with a velocity of 50 m/s . On applying a constant force on ball for 2.0 seconds , it acquires a velocity of 70 m/s .

From this data we can conclude that ;

• Mass of a ball (m) = 10 grams

• Initial velocity of a ball (u) = 50 m/s

• Time taken (t) = 2 seconds

• Final velocity of a ball (v) = 70 m/s

Now,

Let's convert Mass of the ball from grams to kgs

This implies ;

1 gm = 1/1000 kg

10 grams = ?

=> 10/1000

=> 1/100

=> 0.01 kg

Hence,

• Mass of a ball (m) = 0.01 kg

Now,

Let's find the initial momentum of the ball ?

Using the formula ;

• p = mu

=> p = 0.01 x 50

=> p = 0.5 kg.m/s

Hence,

Initial momentum of the ball (mu) = 0.5 kg.m/s

Now,

Let's find the final momentum of the ball ?

Using the formula ;

• p = mv

=> p = 0.01 x 70

=> p = 0.7 kg.m/s

Hence,

Final momentum of the ball (mv) = 0.7 kg.m/s

Now,

Let's find the rate of change of momentum ?

Using the formula ;

• ∆p = (mv - mu)/t

=> ∆p = (0.7 - 0.5)/2

=> ∆p = 0.2/2

=> ∆p = 0.1 kg . m/s²

Hence,

Rate of change of momentum (∆p) = 0.1 kg.m/s²

Now,

Let's find the acceleration of the ball ?

Using the equation of motion ;

v = u + at

=> 70 = 50 + a x 2

=> 70 = 50 + 2a

=> 70 - 50 = 2a

=> 20 = 2a

=> 2a = 20

=> a = 20/2

=> a = 10 m/s²

Hence,

Acceleration of a ball (a) = 10 m/s²

Now,

Let's find the magnitude of the force applied ?

Using the formula ;

• F = ma

=> F = 0.01 x 10

=> F = 0.1 kg.m/s²

=> ( since, kg.m/s² = newton [N] )

=> F = 0.1 N

Hence,

Magnitude of the force applied = 0.1 N

Answer 2

Given:-

⠀

$\sf a \: ball \: of \: mass \: 10g \: is \: moving \: with \: a \: velocity \\ \sf of \: 50m \: {s}^{ - 1} \: on \: applying \: a \: constant \: force \: on \\ \sf a \: ball \: for \: 2 - 0 \: s \: it \: acquires \: a \: velocity \: of \: \: \: \: \\ \sf70 \: m \: {s}^{ - 1} calculate : \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

⠀

To Find:-

⠀

• The initial momentum of ball

• The final momentum of ball

• The rate of change of momentum

• The acceleration of ball

• The magnitude of force applied

⠀

STEP BY STEP EXPLANATION:-

⠀

$\sf m = 10g \: \: \: \: \: \: u = 50 \: \frac{m}{sec} \: \: \: \: \: \: \sqrt{} = 70 \frac{m}{sec} \: \: \: \: \: \: t = 2sec \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf 1) \: initial \: momeutum = m \times u = \frac{10}{1000} \times 50 = 0.5kg \: \frac{m}{sec} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf 2) \: final \: momeutum = m × √ = \frac{10}{1000} \times 70 = \frac{7}{10} = 0.7kg \: \frac{m}{sec} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf3) \: \frac{(final \: momentum - intial \: momentum ) }{t} = \frac{ ( 0.7 - 0.5 )} { \: \: \: 2 = 0.1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf4) \: acceleration = a = \frac{(v - u)}{t} = \frac{20}{2} = 10 \frac{m}{ {sec}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf 5)force = f = m \times a = \frac{10}{1000} \times 10 = 0.1n \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$