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