Question

A bullet of mass 0.03 kg and moving with a velocity of 'x' hits a target with a force of 187.5 November. If the bullet penetrates 0.80m in the target, find the value of x

Answer 1

Answer:

value of x is 100 m/s

Explanation:

1. as ultimately bullet stops so we can consider final velocity as zero m/s.
2. mass of bullet is given 0.03 kg
3. force experienced by bullet is (-)187.5N.    so to calculate deacceleration felt by bullet using     F=m*a
4. a= F/m , substituting values     a = -187.5/ 0.03
5. we'll get a = -6250n
6. using equation of motion
7. $v^2 - u^2 = 2*a*s\\o-x^2= 2*(-6250 N/m)*(0.8m)\\x^2 = 2*5000(m/s)^2\\taking square on both side to get x\\\sqrt{x^2} = \sqrt{10000*(m/s)^2} \\x= 100 m/s$