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Question
AnswersTwo balls A and B are thrown simultaneously, A vertically upwards with a speed of 20m/s from the ground and B vertically downwards from a height of \[40m\] with the same speed and along the same line of motion. At what point do the balls collide? $(g = 9.8m/{s^2})$
Answer
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Hint: Speed of both the balls that were thrown vertically upwards and downwards so the speed and acceleration are the same for both the balls. Hence by applying the second equation of motion we get the answer
Complete step by step answer:
Let us consider the speed of the ball thrown upwards as $U_1$ and the ball thrown downwards as $U_2$.
Then, $U_1 = 20m/s$ and $U_2 = 20m/s$
Consider X as the distance from the ground where the collision takes place then,
Distance from top of the tower up to the collision point = \[40{\text{ }}-{\text{ }}X\]
If it's suppose that collision took place after time T then
For upward motion, $X = 20T - \dfrac{1}{2} \times g \times {T^2}$ (Since the second equation of motion $S = ut + \dfrac{1}{2}a{t^2}$ )
For downward motion, $40 - X = 20T + \dfrac{1}{2} \times g \times {T^2}$
Adding both the equations, we get $40 = 40T$.
Which gives T = 1second
We can substitute the value of T in any of the two equations to get the answer.
Thus by substituting the value of T in the upward motion equation we get,
$X = 20(1) - \dfrac{1}{2} \times 9.8 \times {(1)^2}$ (Taking$(g = 9.8m/{s^2})$)
$X = 20 - 4.9 = 15.1\text{metre}$
So collision occurs at \[15.1\]metres from ground or \[40-15.1 = 24.9\] metres from top.
Note: Studying reference frames shall make these questions exponentially easier.
Assuming the reference frame of the ball A, From this reference frame, since Ball B and Ball A have the same acceleration ’g’ we need not to account for the acceleration.
For the balls to meet, the two must've covered a combined total of \[40\] metres, but from our Reference Frame, ball B must've covered these \[40\]metres with a speed in \[\left({20} + {20} \right)m/s\], or in $\dfrac{{40}}{{40}}$, or in 1 second. Hence by applying time as done in the solution we can find the collision distance.
Complete step by step answer:
Let us consider the speed of the ball thrown upwards as $U_1$ and the ball thrown downwards as $U_2$.
Then, $U_1 = 20m/s$ and $U_2 = 20m/s$
Consider X as the distance from the ground where the collision takes place then,
Distance from top of the tower up to the collision point = \[40{\text{ }}-{\text{ }}X\]
If it's suppose that collision took place after time T then
For upward motion, $X = 20T - \dfrac{1}{2} \times g \times {T^2}$ (Since the second equation of motion $S = ut + \dfrac{1}{2}a{t^2}$ )
For downward motion, $40 - X = 20T + \dfrac{1}{2} \times g \times {T^2}$
Adding both the equations, we get $40 = 40T$.
Which gives T = 1second
We can substitute the value of T in any of the two equations to get the answer.
Thus by substituting the value of T in the upward motion equation we get,
$X = 20(1) - \dfrac{1}{2} \times 9.8 \times {(1)^2}$ (Taking$(g = 9.8m/{s^2})$)
$X = 20 - 4.9 = 15.1\text{metre}$
So collision occurs at \[15.1\]metres from ground or \[40-15.1 = 24.9\] metres from top.
Note: Studying reference frames shall make these questions exponentially easier.
Assuming the reference frame of the ball A, From this reference frame, since Ball B and Ball A have the same acceleration ’g’ we need not to account for the acceleration.
For the balls to meet, the two must've covered a combined total of \[40\] metres, but from our Reference Frame, ball B must've covered these \[40\]metres with a speed in \[\left({20} + {20} \right)m/s\], or in $\dfrac{{40}}{{40}}$, or in 1 second. Hence by applying time as done in the solution we can find the collision distance.
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