# Will a Ping-Pong Ball Hit a Doll?

• courtney1121
In summary, the question of whether the ping-pong ball will hit the doll depends on how you define "aiming at the doll." If it means pointing the gun so that the initial velocity of the projectile is directly at the doll, then the ball will likely hit the doll. However, if it means pointing the gun so that the doll's current position is in the projectile's flight path, the ball may not hit the doll. This can be determined by using equations for the height and horizontal position of the ball, as well as the height of the doll, to compare their respective heights after the ball has traveled a horizontal distance.
courtney1121
Consider the following: You have a toy gun that shoots ping-pong balls at about 5 m/s. You aim the gun at a doll sitting on a fence. Just as you pull the trigger, a friend of yours (or so you thought) nudged the doll forward, just enough to fall off the fence. Will the ping-pong ball hit the doll? Why or why not?

I'm confused on where to start. So I probably need to assume a distance, correct? Also, do I have to treat the two objects as different projectiles, or just worry about the ping-pong ball?

The distance does not matter, as long as the ground does not ineterfere. It turns out the speed of the ball does not matter either. You can just call the distance d and the speed v_o

Write the equations for the height and the horizontal position of the ball, using initial velocities that correspond to aiming at the doll. Write the equation for the height of the doll. Compare the height of the doll to the height of the ball after the ball has moved a horizontal distance d.

so one is in the x direction and the other is in the y direction. ball is in the x and the doll is the y direction.

x = x0 + v0xt
y = y0 +v0yt - 1/2gt^2

or am I looking at this wrong?

or actually height would be v0t+(.5)at^2...?

courtney1121 said:
so one is in the x direction and the other is in the y direction. ball is in the x and the doll is the y direction.

x = x0 + v0xt
y = y0 +v0yt - 1/2gt^2

or am I looking at this wrong?

The ball moves in both directions. The ball is aimed at the doll, so its initial velocity is resolved into horizontal and vertical components.

courtney1121 said:
or actually height would be v0t+(.5)at^2...?

That's good for the ball. You need the one with y0 for the doll.

Last edited:
y = 1/2gt^2?

I found the time to be .5s because v/g so the height for the ball is 5*.5 + (.5)*9.8*(.5)^2 which is 6.225m.

courtney1121 said:
y = 1/2gt^2?

For the doll, you want the equation you wrote before

y = y0 +v0yt - 1/2gt^2

but recognize that it starts at rest at some initial height h above the starting point of the ball. You will want the same equation for the ball, but starting from a different intial height (perhaps a good place to call zero) with an initial velocity.

courtney1121 said:
I found the time to be .5s because v/g so the height for the ball is 5*.5 + (.5)*9.8*(.5)^2 which is 6.225m.

How could you find a time? There is no specific distance given from the gun to the doll.

courtney1121 said:
Consider the following: You have a toy gun that shoots ping-pong balls at about 5 m/s. You aim the gun at a doll sitting on a fence. Just as you pull the trigger, a friend of yours (or so you thought) nudged the doll forward, just enough to fall off the fence. Will the ping-pong ball hit the doll? Why or why not?

Well, the answer depends a bit on what is meant by aiming at the doll. If aiming at the doll means pointing the gun so that the inital velocity of the projectile is directly at the doll, you'll get one answer. If aiming at the doll means pointing the gun so that the current position of the doll is in the projectile's flight path, you might get a different one.

## 1. Will a Ping-Pong Ball Hit a Doll?

It depends on the distance between the Ping-Pong ball and the doll, the speed of the Ping-Pong ball, and the size and shape of the doll. Generally, if the Ping-Pong ball is close enough and traveling at a high enough velocity, it will hit the doll.

## 2. What happens when a Ping-Pong Ball hits a doll?

The Ping-Pong ball will transfer its kinetic energy to the doll, causing the doll to move in the direction of the ball's impact. Depending on the force of the impact, the doll may also experience some deformation or damage.

## 3. Can a Ping-Pong Ball break a doll?

It is unlikely that a Ping-Pong ball alone would be able to break a doll, as it does not have enough mass or force. However, if the Ping-Pong ball is launched at high speeds or combined with other objects, it may be able to cause damage to the doll.

## 4. What factors affect the likelihood of a Ping-Pong Ball hitting a doll?

The main factors that affect the likelihood of a Ping-Pong ball hitting a doll are distance, velocity, and the size and shape of the doll. Other factors such as air resistance, gravity, and the surface the doll is on may also play a role.

## 5. Are there any safety precautions to consider when conducting experiments with Ping-Pong balls and dolls?

Yes, it is important to wear safety goggles when conducting experiments with Ping-Pong balls and dolls, as the balls may bounce or ricochet in unexpected directions. It is also important to ensure that the experiment is conducted in a safe and controlled environment to prevent any accidents or injuries.

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