4. A 20 gram bullet moving at 300 m/s stops afterpenetrating 3 cm of b

1.	4. A 20 gram bullet moving at 300 m/s stops afterpenetrating 3 cm of bone. Calculate the averageforce exerted by the bullet. [Ans. â€“ 3 x 104 N]
Answer» There are two methods to solve this. METHOD 1 : Using only Motion Equations. m = 0.02 kgu = 300 m/sv = 0d = 0.03 m(Note everything is in SI system of units) a = (v×v - u×u)/2s = (0-900)/0.06 = -15000 m/s^2- sign indicates that acceleration is opposite to motion. This type of acceleration is also called retardation or deceleration. So, Force = ma = 0.02 × 15000 = 300 newton (Ans) METHOD 2 : Using Conservation of Energy. So, we have Work done = Change in Kinetic EnergyHence, Fd = mv×v/2F = mv×v/(2d) = 0.02 × 30 × 30 / 0.06 = 300 newton (Ans) work done, fd=mvxv/2d==0.0××30×30/0.06=300 newton 300 Newton is the answer

4. A 20 gram bullet moving at 300 m/s stops afterpenetrating 3 cm of bone. Calculate the averageforce exerted by the bullet. [Ans. â€“ 3 x 104 N]

Answer»

There are two methods to solve this.

METHOD 1 : Using only Motion Equations.

m = 0.02 kgu = 300 m/sv = 0d = 0.03 m(Note everything is in SI system of units)

a = (v×v - u×u)/2s = (0-900)/0.06 = -15000 m/s^2- sign indicates that acceleration is opposite to motion. This type of acceleration is also called retardation or deceleration.

So, Force = ma = 0.02 × 15000 = 300 newton (Ans)

METHOD 2 : Using Conservation of Energy.

So, we have Work done = Change in Kinetic EnergyHence, Fd = mv×v/2F = mv×v/(2d) = 0.02 × 30 × 30 / 0.06 = 300 newton (Ans)

work done, fd=mvxv/2d==0.0××30×30/0.06=300 newton

300 Newton is the answer

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