# A lead bullet of mass 0.05kg is fired with a velocity of 200m/s…

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**A lead bullet of mass 0.05kg is fired with a velocity of 200m/s into a lead block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is**

**A. 150 J**

**B. 100 J**

**C. 50 J**

**D. 200 J**

**QUICK ANSWER…**

**C**

**DETAILS…**

**Please note that if the lead block can move freely, then the momentum of the bullet will be conserved after impact. This also means the momentum of the bullet will be transferred into the block.**

**Hence: total momentum before impact = total momentum after impact**

**please note that the bullet entered and became one with the block after impact.**

**(Mass of bullet x velocity of bullet) + (Mass of block x velocity of block) = (mass of block and bullet (together) x velocity of block and bullet (together))**

**Data given:**

**mass of bullet = 0.05kg**

**velocity of bullet (before impact) = 200m/s**

**mass of block = 0.95kg**

**velocity of block (before impact) = 0**

**0.05 x 200 + 0.95 x 0 = (0.05 + 0.95) x velocity of block and bullet.**

**10 + 0 = (1) x velocity of block and bullet.**

**Therefore, velocity of block and bullet together = 10m/s**

**Final kinetic energy after impact will be = ½ mv^2 = ½ x (mass of block and bullet) x (square of velocity of block and bullet)**

**= ½ x (0.05 +0.95) x (10 x 10) = 0.5 x 1 x 100 = 50Joules**

**Now for the right answer to the above question:**

**Option A is incorrect.****Option B is incorrect.****C is correct.****D is not the correct answer.**

**HELPFUL TIPS…**

**You may please note these/this:**

**In this question, it is assumed that the bullet entered into the block and became one with it, transferring its momentum into it.**

**Use the questions and answers session to correct, express, and contribute your views.**

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**/ culled from 2018 JAMB-UTME physics question 24 /**