Question

a bullt of 0.05 kg strikes a wooden target 0.2 m thick with a velocity of 300 m/s and emerges out with a velocity of 100m/s. calculate the:- (1) loss in kinetic energy and (2) average force of friction of wood
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Answer 1

Given :

• Mass of the bullet = 0.05 kg
• Thickness of the wood (d) = 0.2 m
• Initial velocity (u) = 300 m/s
• Final velocity (v) = 100 m/s

To find :

• Loss in KE
• Average force of friction acting on the wood .

Solution :

(1) Kinetic energy is the energy possesed by the body due to it's motion .

⇒ KE = ½ mv²

Initial kinetic energy

⇒ KE = ½ m u²

Now let's substitute the given values in the above equation ,

⇒ KE = ½ x 0.05  x 90,000

⇒ KE = 4,500 J

Final kinetic energy

⇒ KE' = ½ m v²

Now let's substitute the given values in the above equation ,

⇒ KE' = ½ x 0.05 x 10000

⇒ KE = 500 J

Loss  in kinetic energy

$\implies \mathtt{\Delta KE = KE' - KE}$

⇒ ΔKE = 4,500 - 500

⇒ ΔKE = 4000 J

The increase in kinetic energy is 4000 J .

(2)

⇒  Work done = ΔKE  

 

We know that

⇒ Work done = F. d

Hence ,

⇒ ΔKE  = F. d

⇒ F = 4000 / 0.2

⇒ F = 20,000 N

The average force of friction acting on the wood is 20,000 N

Answer 2

(i) loss in K.E = 2000 J

(ii) avg. force of friction = 10, 000 N

Explanation:

Given,

mass of of bullet = 0.05 kg

thickness of wooden target = 0.2 m

initial velocity (u) = 300 ms‐¹

final velocity (v) = 100 ms‐¹

Initial K.E = 1/2mu²

= 1/2×0.05×300×300

= 2250 J

Final K.E = 1/2mv²

= 1/2×0.05×100×100

= 250 J

(i) Loss in K.E = Initial K.E - Final K.E

= 2250 J - 250 J

= 2000 J

(ii) Work done by the bullet = average force of friction of wood × distance traveled by the bullet

W = F × S

2000 = F × 0.2

F = 2000 ÷ 0.2

F = 10, 000 N

Hence, average force of friction of wood = 10, 000 N.