I won't possibly throw like that, would I?
Ball A of mass 1 kg is thrown at an angle of 4 5 ∘ with the horizontal with kinetic energy 5 0 J such that it hits ball B of the same mass placed on the top of a pole. In this collision, half of the kinetic energy of ball A at that instant is transferred to ball B , causing ball B to move in a forward direction. If it is known that the height at which ball B was placed is the maximum height that the initial projectile of ball A would have traveled, then find the distance of the final position of ball B from the foot of the pole.
The figure below would help in understanding the situation:
Details and Assumptions:
- Ball B does not rebound after hitting the ground.
- Air friction is negligible.
- Take g = 1 0 m/s 2 as the acceleration due to gravity.
- Give your answer (in m ) to two decimal places.
The answer is 3.54.
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1 solution
This is a fantastic problem. Simple and yet challenging
I missed the answer, but would like to give my method below.
Since the angle is 45 degrees, Sin45=Cos45, and the Vertical and Horizontal velocities, and so K.E. in these directions will also be equal.
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.
Note:- After collision, KE is half, half. Momentum before and after is NOT the same.
I've done exactly the same way. Can this problem be solved in other ways?
Same solution
i don't think it deserved to be at level 3 just a problem of level 1. wasted my time
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Well, it takes so much time to come up with such a problem. Atleast i am contributing. Happy to see it got 8 likes. Well, criticisms are a part of it. :) Talking about the level, it automatically changes with the number of people solving it right, you dont need to worry about it.
Is'nt this question is overated!!
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u should read my comment at that time this was level 3....
kinetic not kibetic but otherwise great solution!
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Hehe thanks for mentioning the spelling mistake :)
Kinetic energy of ball A = 5 0 J .
Now,
K.E. 5 0 v 2 v = 2 1 mv 2 = 2 1 × 1 × v 2 = 1 0 0 = 1 0 m/s
At the maximum point of trajectory, the vertical component of velocity of ball A becomes zero and what remains is the horizontal component of ball A which is given by v cos θ = 1 0 cos 4 5 ° = 2 1 0 .
Kinetic energy at this point = 2 1 mv 2 cos 2 θ = 2 1 × 1 × 2 1 0 0 = 2 5 J
So, kinetic energy obtained by ball B = 2 2 5 . (Ball B obtains half the kinetic energy of ball A).
Again using the formula,
K.E. 2 2 5 v 2 v = 2 1 mv 2 = 2 1 × 1 × v 2 = 2 5 = 5 m/s .
The height of the pole is the maximum height of trajectory of ball A which is given by 2 g v 2 sin 2 θ = 2 × 1 0 1 0 0 × 2 1 = 2 . 5 m
Since ball B is falling from a height, range of its projectile is given by u g 2 h = 5 1 0 2 × 2 . 5 = 5 2 1 = 2 5 ≈ 3 . 5 4 m