Question

9. A bullet of mass 10g strikes a wooden plank at a speed of 500m/s and gets embedded after travelling 5cm. Calculate (i) the resistive force exerted by the wooden plank on the bullet (ii) the time taken by the bullet to come to rest.​

Answer 1

Answer:

see the attachments given.....

HOPE THIS HELPS YOU.....✌️✌️

Answer 2

Answer:

i) F = 25 kN = 25,000 N

ii) t = 2 * 10⁻⁴ seconds

Explanation:

Given data:

m = 10 g = 10/1000 = 0.01 kg

v = 0 m/sec

s = 5 cm = 0.05 m

u = 500 m/s

v² = u² + 2*a*s

a = 500² / (2*0.05) = 2.5 * 10⁶ m/sec² (decelration)

According to Newton's laws of motion we know,

F = m * a

where F is in Newtons (N)

m is mass in kilogram (kg)

and a is acceleration in m/sec².

Coming to the question, we know

a = 2.5 * 10⁶ m/sec² and m = 0.01 kg

put in above equation,

F = 0.01 * 2.5 * 10⁶

F = 25 kN resistive force.

Time taken to stop:

v = u + a*t

t = -500 / -2.5 * 10⁶ = 2 * 10⁻⁴ seconds