Question

A bullet of mass 25 g moving with a speed of 100m/s pierces a bag full of sand kept adjacent to a wall.The bullet stops in the bag after 0.05 second.Find a)acceleration of the bullet b)The force exerted by sandbag on the bullet c)the distance covered by bullet before coming to test THE WRONG ANSWER WILL BE REPORTED AND THE RIGHT ANSWER WILL BE MARKED AS BRAINLIST​

Answer 1

Given that, a bullet of mass 'm' 25 g moving with a speed of 100 m/s. Also the bullet stops in the bag after 0.05 second.

We have to find the -

a)acceleration of the bullet

b)The force exerted by sandbag on the bullet

c)the distance covered by bullet before coming to test.

From above data, we have u = 100 m/s, m = 25 g = 0.025 kg, v = 0 m/s and t = 0.05 second

(a) Using the First Equation Of Motion,

v = u + at

Substitute the known values

→ 0 = 100 + a(0.05)

→ -100 = 0.05a

→ -10000/5 = a

→ -2000 = a (negative sign shows retardation)

Therefore, the acceleration of the bullet is -2000 m/s².

(b) Now, force is defined as the product of mass and acceleration.

F = ma

Substitute the values,

→ F = 0.025 × (-2000)

→ F = -50 N

Therefore, the force on the bullet is -50N.

(c) Using the Second Equation Of Motion,

s = ut + 1/2at²

Substitute the values

→ s = 100(0.05) + 1/2 × (-2000) × (0.05)²

→ s = 5 - 1000(0.0025)

→ s = 5 - 2.5

→ s = 2.5

Therefore, the distance covered by the bullet before coming to test is 2.5 m.

Answer 2

Given:- Mass of a bullet (m)=25g[convert it into kg= 0.025kg,Speed of the bullet=100m/s,Time when bullet stops (t)=0.05sec

Solution:

a.)We have to find the acceleration of the bullet.

We know acceleration (a)= v-u/t

=0-100/0.05=-2000m/s²

b.) Force exerted by the sandbag on the bullet

→Force=m×a

→F=0.025×(-2000)

→F=-50N

c.) The distance covered by the bullet

→S= ut+½at²

→S=100×0.05+½×(-2000)×(0.05)²

→S=2.5m

Hence,the acceleration is -2000m/s², Force is -50N and Distance is 2.5N