A bullet of mass 4g fired with a velocity of 50m/s enters a wall up to a depth of 10cm . How will the depth of penetration into the wall change if a bullet of mass 5g strikes against it with a velocity of 40ms-1 ? give reason to justify your answer.
Answers
Deceleration of the bullet = a = (0² - u1²)/ (2*s1) = (0 - 50²)/0.2
a = - 12500 m/s²
Resistive Force = m a = - 50 N
We assume that the bullets are nearly of the same surface area on their front side and of same shape. The wall is uniform in its resistive quality. The pressure offered by the wall remains same. So impact resistive force offered by wall is same in both cases. Because Pressure * surface area is same. If the 5 g bullet is bigger (front surface area) then pressure will be more.
m2 = 5 g.
deceleration = a2 = - 50 N / 0.005 = - 10,000 m/sec²
distance of penetration = s2 = (0 - v2²) /2 (a2)
= - 40² / [2 * -10000] m
= 0.08 m or 8 cm
Answer:
m1 = 4g u1 = 50 m/s s1 = penetration into wall = 10 cm
Deceleration of the bullet = a = (0² - u1²)/ (2*s1) = (0 - 50²)/0.2
a = - 12500 m/s²
Resistive Force = m a = - 50 N
We assume that the bullets are nearly of the same surface area on their front side and of same shape. The wall is uniform in its resistive quality. The pressure offered by the wall remains same. So impact resistive force offered by wall is same in both cases. Because Pressure * surface area is same. If the 5 g bullet is bigger (front surface area) then pressure will be more.
m2 = 5 g.
deceleration = a2 = - 50 N / 0.005 = - 10,000 m/sec²
distance of penetration = s2 = (0 - v2²) /2 (a2)
= - 40² / [2 * -10000] m
= 0.08 m or 8 cm