Problem based on Force and Law of motion

1. A bullet of mass 12 gram hits a target with 50 m/sec and stop in 0.02 sec.

a). The distance covered by bullet before stoping.

b). Force exerted by target on bullet.

Sol. Given m = 12 gram = 0.012 Kg u = 50 m/sec v = 0 t = 0.02 sec

a). v = u + a t

0 = 50 + a * 0.02

- a * 0.02 = 50

a = -50/0.02

a = -2500 m/sec^2

v^2 = u^2 + 2as

0 = (50)^2 + 2 (-2500) * s

5000 s = 50 * 50

S = 2500 / 5000

S = 1 / 2 = 0.5 m

b). f = m a

= 0.012 * 2500 = 30 N

2. A bullet of mass 15 gram hits a target with 150 m/sec and stop in 0.04 sec.

a). The distance covered by bullet before stoping.

b). Force exerted by target on bullet.

Sol. Given m = 15 gram = 0.015 Kg u = 150 m/sec v = 0 t = 0.04 sec

a). v = u + a t

0 = 150 + a * 0.04

- a * 0.04 = 150

a = -150/0.04

a = -3750 m/sec^2

v^2 = u^2 + 2as

0 = (150)^2 + 2 (-3750) * s

7500 s = 150 * 150

S = 22500 / 7500

S = 3 m

b). f = m a

= 0.015 * 3750 = 56.25 N

3. If a car is moving with initial velocity of 72 km/h and sudden applying break and velocity decreases to 36 km/h in 10 sec. Find the acceleration of car.

Sol. Given t = 10 sec u = 72 km/h = (72 * 5) / 18 = 20 m/sec

v = 36 km/h = (36*5) / 18 = 10 m/sec

v = u + a t

a = ( v – u ) / t

= ( 20 – 10 ) / 10

= ( 10 / 10 ) = 1 m/sec^2

4. If a car is moving with initial velocity of 36 km/h and sudden applying break and velocity decreases to 72 km/h in 20 sec. Find the acceleration of car.

Sol. Given t = 20 sec v = 72 km/h = (72 * 5) / 18 = 20 m/sec

u = 36 km/h = (36*5) / 18 = 10 m/sec

v = u + a t

a = ( v – u ) / t

= ( 10 – 20 ) / 20

= ( -10 / 20 ) = ½ = 0.5 m/sec^2

5. A body of mass 100 Kg start moving from rest and covered a distance of 1600 mm. with a acceleration of 20 m/ sec^2. Find final velocity of body and also find the momentum of body.

Sol. Given m = 100 kg s = 1600 mm = 1.6 m

a = 20 m/sec^2 u = 0 v = ?

v^2 = u^2 + 2 a s

= 0 + 2 * 20 * 1.6

V^2 = 64 = 8^2

V = 8 m/sec

P = m v

= 100 * 8

= 800 kg m /sec

6. A ball of mass 500g moving with velocity of 20 m/sec and hits another ball of mass 4kg (rest). Find the combined velocity of balls after collision.

Sol. Given m1 = 500g = 0.5 kg m2 = 4 kg v1 = 20 m/sec v2 = 0

Total momentum before collision = ( m1 v1 + m2 v2 )

= ( 0.5 * 20 + 4 * 0 )

= 10 kg m/sec

Total momentum before collision = Total momentum after collision

( m1 v1 + m2 v2 ) = ( m1 + m2 ) v

V = ( m1 v1 + m2 v2 ) / ( m1 + m2 )

= ( 0.5 * 20 + 4 * 0 ) / ( 0.5 + 4 )

= 10 / 4.5

V = 2.22 m/sec

science-ncert-solution

7. A car of mass 100kg moving with velocity of 20 m/sec for 20 sec and hits tree wall. Find the distance covered by car and also find the impact force.

Sol. Given m = 100kg u = 20 m/sec t = 20 sec v = 0

V = u + a t

a = ( v – u ) / t

= ( 0 – 20 ) / 20

= -1 m/sec^2

V^2 = u^2 + 2 a s

0 = 20^2 + 2 (-1) s

2 s = 400

S = 200 m

F = m a

= 100 * (-1)

= -100 N

Milan Tomic

Hi. I’m Designer of Blog Magic. I’m CEO/Founder of ThemeXpose. I’m Creative Art Director, Web Designer, UI/UX Designer, Interaction Designer, Industrial Designer, Web Developer, Business Enthusiast, StartUp Enthusiast, Speaker, Writer and Photographer. Inspired to make things looks better.