Question

A ball of mass 0.2 kg rests on a vertical of heights 5 m. a bullet of mass 0.01 kg, travelling with a velocity v m/s in a horizontal direction, hits the center of the ball. after the collision, the ball and bullet travel independently. they hit the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. the initial velocity v of the bullet is____________.

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Answer 1

Hey my dear friend arivu
We will use conservation of momentum in the end 
mv + MV = mv' + MV' 
where the non primed velocities are before the collision and the primed velocities are after the collision. 
Lets find the primed velocities from the distances traveled after the collision. 
The time it takes each object to fall 5 m is the same 
5 = 0.5gt 
t = 1.02sec. 
Now to find the velocities 
20 = V'(1.02) 
V' = 19.6 m/s 
100 = v'(1.02) 
v' = 98 m/s 
Now going back to the momentum equation 
0.01v + 0.2(0) = 0.01(98) + 0.2(19.6) 
0.01v = 4.9 
v = 490 m/s (so I'd go with (D) 500 m/s)