A bullet of mass 'm' moving with a speed 'v' strikes a wooden block of mass 4m' at rest and gets embedded in it. Then the speed of this
embedded block will be (assume surface on which block is lying is frictionless):
O 5v
0 v/5
0 4v
0 v/4
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Answer:
v/4
Explanation:
by using law of conservation of momentum
(mass of bullet × velocity of bullet) +(mass of block× velocity of block)(before collision)=(mass of bullet×velocity of bullet)+(mass of block × velocity of block) (after collision)
mv+4m×0= m×0+4m×a-(let a velocity of block)
mv+0=0+4ma
mv= 4ma
a= mv/4m
a = v/ 4
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