A bullet of 10 g moving with a speed of 100 m/s penetrates a sandbag
and comes to rest in 1/10 th second Find
a) The distance through which the bullet penetrates
b) The retarding force experienced
Answers
Answer:
Initial velocity, u= 150 m/s
Final velocity, v= 0 (since the bullet finally comes to rest)
Time taken to come to rest, t= 0.03 s
According to the first equation of motion, v= u + at
Acceleration of the bullet, a
0 = 150 + (a — 0.03 s)a = -150 / 0.03 = -5000 m/s2
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:
v2= u2+ 2as
0 = (150)2+ 2 (-5000)
= 22500 / 10000
= 2.25 m
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass * Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s2
F = ma = 0.01 * 5000 = 50 N
Hence, the magnitude of the force exerted by the wooden block on the bullet is 50 N.
The answers are as follows;
The answers are as follows; (a) -10N
The answers are as follows; (a) -10N (b) 5 meters
GIVEN
A bullet of 10 g moving with a speed of 100 m/s penetrates a sandbag and comes to rest in 1/10 th second.
TO FIND
a) The distance through which the bullet penetrates
b) The retarding force experienced.
SOLUTION
We can simply solve the above problem as follows;
It is given,
Mass of the bullet = 10 grams = 0.01 kg
Initial velocity of bullet = 100m/s
Time taken = 1/10 = 0.1 second
Final velocity = 0
Applying the formula to calculate acceleration
v = u+at
0 = 100 + 0.01a
a = - 100/0.01 = - 1000m/s²
To calculate the retarding force, We will apply the following formula,
F = ma
Putting the values in the above formula we get,
F = 0.01 × - 1000m/s²
= -10N
Applying second equation of motion to calculate distance;
=
= 10-5= 5 meters.
Hence, The answers are as follows;
- Hence, The answers are as follows; (a) -10N
- Hence, The answers are as follows; (a) -10N (b) 5 meters
#spj2