Question

Question 14 A bullet of mass 10 g travelling horizontally with a velocity of 150 m s −1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Class 9 - Science - Force and Laws of Motion Page 129

Answer 1

Data:
mass of the bullet (m) = 10 g = (10/1000) kg =0.01 kg
Initial velocity (u) = 150 m/s
Final velocity (v) = 0 m/s
Time taken (t) = 0.03 sec

i. Calculation of the distance of penetrationof the bullet into the block:  

Calculation of the distance of penetrationof the bullet into the block = Displacement

Displacement = average velocity × time
Displacement = [(u + v) / 2] × time
Displacement = [(150 + 0) / 2] × 0.03
Displacement = 2.25 m


ii. Calculation of the magnitude of the force exerted by the wooden block on the bullet:
the magnitude of the force exerted by the wooden block on the bullet = Retarding force
Retarding force = mass × reatardation
Retarding force = mass ×[ (Intitial velocity - Final velocity) / t ]
Retarding force = 0.01 ×[ (150 - 0) / 0.03 ] N
Retarding force = [150/3] N
Retarding force = 50 N

Answer 2

Mass of bullet , (m) = 10g = 0.01 kg

Initial velocity of bullet , u = 150 ms-¹

Final velocity of bullet , v = 0

Time Taken , t = 0.03 s.


◆ ¡) using , v = u + at

or , a = v - u/t

= ( 0- 150) ms-¹/ 0.03 s

= - 5000 ms-²


Using, S = ut + 1/2at² , we get

S = 150 × 0.03 - 1/2 × 5000 × 0.03

= 4.5 - 2.25

= 2.25 m

Therefore, Distance of penetration of bullet into the block = 2.25 m.


◆ ii) Using, F = ma, we get

F = 0.01 × (-5000)

= -50N

Therefore, Magnitude of force = 50N