Two balls, A and B, are thrown simultaneously, a vertically upward with a speed of 20 m/s from the ground and B vertically downward from a height of 40 m with the same speed and along the same line of motion. At what points do the two balls collide by taking acceleration due to gravity as 9.8 m/s ²?

Question

Prepp Team · Accepted Answer

Understanding the Vertical Motion and Collision

This physics problem involves two objects (balls A and B) moving under the influence of gravity along the same vertical line. Ball A is thrown upwards from the ground, while Ball B is thrown downwards from a certain height. We need to find when and where they collide.

Let's define our coordinate system. We will take the ground as the origin ($y=0$) and the upward direction as positive.

Setting Up Equations of Motion

We can use the standard kinematic equation for displacement under constant acceleration:

$ y = y_0 + v_0t + \frac{1}{2}at^2 $

Here, $ y $ is the final position, $ y_0 $ is the initial position, $ v_0 $ is the initial velocity, $ a $ is the acceleration, and $ t $ is the time.

Motion of Ball A (Upward from Ground)

Initial position, $ y_{A0} = 0 $ m (from the ground).
Initial velocity, $ v_{A0} = +20 $ m/s (upward is positive).
Acceleration due to gravity, $ a_A = -g = -9.8 $ m/s${^2}$ (acting downwards).

The position of Ball A at time $ t $ is:

$ y_A(t) = y_{A0} + v_{A0}t + \frac{1}{2}a_At^2 $

$ y_A(t) = 0 + (20)t + \frac{1}{2}(-9.8)t^2 $

$ y_A(t) = 20t - 4.9t^2 $

Motion of Ball B (Downward from 40 m Height)

Initial position, $ y_{B0} = 40 $ m (from the ground).
Initial velocity, $ v_{B0} = -20 $ m/s (downward is negative).
Acceleration due to gravity, $ a_B = -g = -9.8 $ m/s${^2}$ (acting downwards).

The position of Ball B at time $ t $ is:

$ y_B(t) = y_{B0} + v_{B0}t + \frac{1}{2}a_Bt^2 $

$ y_B(t) = 40 + (-20)t + \frac{1}{2}(-9.8)t^2 $

$ y_B(t) = 40 - 20t - 4.9t^2 $

Calculating Collision Time

The balls collide when they are at the same vertical position. So, we set $ y_A(t) = y_B(t) $.

$ 20t - 4.9t^2 = 40 - 20t - 4.9t^2 $

Notice that the term $ -4.9t^2 $ appears on both sides of the equation. We can cancel it out:

$ 20t = 40 - 20t $

Now, we solve for $ t $. Add $ 20t $ to both sides:

$ 20t + 20t = 40 $

$ 40t = 40 $

Divide by 40:

$ t = \frac{40}{40} $

$ t = 1 $ second

The two balls collide after 1 second.

Calculating Collision Height

To find the height at which they collide, we substitute the collision time $ t = 1 $ s into either the equation for $ y_A(t) $ or $ y_B(t) $.

Using the equation for Ball A:

$ y_A(1) = 20(1) - 4.9(1)^2 $

$ y_A(1) = 20 - 4.9 $

$ y_A(1) = 15.1 $ meters

Using the equation for Ball B (just to verify):

$ y_B(1) = 40 - 20(1) - 4.9(1)^2 $

$ y_B(1) = 40 - 20 - 4.9 $

$ y_B(1) = 20 - 4.9 $

$ y_B(1) = 15.1 $ meters

Both equations give the same height, as expected. The collision occurs at a height of 15.1 meters from the ground.

Conclusion

The balls collide after 1 second at a height of 15.1 m from the ground.

Comparing this result with the given options, we find that it matches the third option.

Physics Problem Solving: Revision Table

Concept	Description	Key Formula
Vertical Motion	Motion under gravity along the vertical axis.	$ v = v_0 + at $
Displacement	Change in position.	$ y = y_0 + v_0t + \frac{1}{2}at^2 $
Acceleration due to Gravity	Constant acceleration $ g \approx 9.8 $ m/s${^2}$ downwards.	$ a = -g $ (if up is positive)
Collision Condition	When two objects meet at the same position at the same time.	$ y_1(t) = y_2(t) $

Additional Information on Vertical Motion Physics

Problems involving vertical motion under gravity are classic examples of motion with constant acceleration. Here are a few key points:

Constant Acceleration: The acceleration due to gravity ($g$) is considered constant near the Earth's surface and always acts downwards. Its value is approximately 9.8 m/s${^2}$.
Sign Convention: It is crucial to maintain a consistent sign convention for position, velocity, and acceleration. If upward is taken as positive, then downward quantities (like initial velocity if thrown down or gravity's acceleration) must be negative.
Peak Height: For an object thrown upwards, the velocity at the peak of its trajectory is momentarily zero.
Symmetry (in vacuum): If air resistance is ignored, the time taken to reach the peak is equal to the time taken to fall back to the initial height, and the speed upon returning to the initial height is equal to the initial speed.
Simultaneous Motion: When dealing with multiple objects moving simultaneously, writing down the equation of motion for each object separately based on their initial conditions is the first step. The condition for them to meet (like collision) often involves setting their positions equal at the same time $t$.

Understanding these fundamental concepts helps in solving a wide range of physics problems related to projectile motion and vertical kinematics.

Prepp Team · Answer

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