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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