Question

4. Calculate the momentum of a body of mass 450 g moving with a speed of 120 km/h.
5. A bullet of mass 20 g travelling with a velocity of 20 m/s penetrates a sand bag and comes to rest in
0.05 second. Find a) the distance through which it penetrates b) retarding force.​

Answer 1

Answer:

p(momentum)= mv

= 450/1000 ×120

= 54 kgkm/h

Answer 2

4. Calculate the momentum of a body of mass 450 g moving with a speed of 120 km/h.

mass of the body = 450 g

In order to convert g into kg Divide by 1000

mass of the body = 450/1000 = 0.45 kg

________________

velocity of the body = 120 km / hr

In order to convert km/hr into m /s multiply 5 / 18

velocity of the body

= 120 × 5 / 18

= 600/18

= 33.33 m/s

velocity of the body = 33.33 m / s

_____________

momentum of the body is given by the formula ,

= mass × velocity

= 0.45 × 33.33

= 14.9 kg m / s

The momentum of the body is 14.9 kg m/s

__________________________

5. A bullet of mass 20 g travelling with a velocity of 20 m/s penetrates a sand bag and comes to rest in

A bullet of mass 20 g travelling with a velocity of 20 m/s penetrates a sand bag and comes to rest in0.05 second.

Find

(b) retarding force

• initial velocity of the bullet = 20 m / s

• final velocity of the bullet = 0 m / s

• time taken by the bullet to come to rest = 0.05 sec

• mass of the bullet = 20 g = 0.02 kg

In order to find the acceleration of the bullet let us use the first kinematic equation,

$\rightarrow \mathtt{v = u + at}$

here ,

• v = final velocity
• u =initial velocity
• a = acceleration
• t = time taken

$\rightarrow \mathtt{a = \dfrac{v - u}{t} }$

$\rightarrow \mathtt{a = \dfrac{0 - 20}{0.05} }$

$\rightarrow \mathtt{a = \dfrac{ - 2000}{5} }$

$\rightarrow \mathtt{a = - 400 \dfrac{m}{ {s}^{2} } }$

acceleration of the bullet = 400 m / s²

Note : negative sign denotes retardation

_______________

Retarding force of the bullet

= mass × acceleration

= 0.02 × - 400

= - 8 N

Retarding force of the bullet = - 8 N

__________________________

a) the distance through which it penetrates

• initial velocity of the bullet = 20 m / s

• final velocity of the bullet = 0 m / s

• time taken by the bullet to come to rest = 0.05 sec
• acceleration of the bullet = -400 m /s

In order to find the acceleration of the bullet let us use the third kinematic equation,

$\rightarrow \mathtt{ {v}^{2} = {u}^{2} + 2as}$

here ,

• v = final velocity
• u = initial velocity
• a = acceleration
• s = distance

$\rightarrow \mathtt{ {u}^{2} = - 2as}$

$\rightarrow \mathtt{s= \dfrac{ {u}^{2} }{ - a \times 2} }$

$\rightarrow \mathtt{s= \dfrac{ 400}{ + 400 \times 2} }$

$\rightarrow \mathtt{s= 0.5 \: m}$

The distance through which the bullet penetrates is 0.5 m