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Thursday, December 28, 2017

A lead pellet of mass 10.0 g is shot horizontally into a stationary wooden block of mass 100 g. The pellet hits the block with an impact velocity of 250 m s-1.



Question 3
A lead pellet of mass 10.0 g is shot horizontally into a stationary wooden block of mass 100 g. The pellet hits the block with an impact velocity of 250 m s-1. It embeds itself in the block and it does not emerge.



What will be the speed of the block immediately after the pellet is embedded?
A 23 m s-1                          B 25 m s-1                          C 75 m s-1                         D 79 m s-1





Reference: Past Exam Paper – November 2013 Paper 13 Q13





Solution:
Answer: A.

From the conservation of momentum,
Sum of momentum before collision = Sum of momentum of collision


Momentum = mass × velocity = mv

Sum of momentum before collision = (10×10-3 × 250) + (100×10-3 × 0)
Sum of momentum before collision = (10×10-3 × 250)


After collision, the pellet is embedded into the block. Both can now be considered as a single body of (100+10 =) 110 g. Let the speed with which they move = v.

Momentum after collision = (100+10) ×10-3 × v


Sum of momentum before collision = Sum of momentum of collision
(10×10-3 × 250) = (100+10) ×10-3 × v

Speed v = (10 × 250) / 110 = 22.7 = 23 m s-1

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