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kiruha [24]
3 years ago
7

A very light ping-pong ball moving east at a speed of 4 m/s collides with a very heavy stationary bowling ball. The Ping-Pong ba

ll bounces back to the west, and the bowling ball moves very slowly to the east. Which object experiences the greater magnitude impulse during the collision?A very light ping-pong ball moving east at a speed of 4 m/s collides with a very heavy stationary bowling ball. The Ping-Pong ball bounces back to the west, and the bowling ball moves very slowly to the east. Which object experiences the greater magnitude impulse during the collision?
Physics
1 answer:
KengaRu [80]3 years ago
3 0

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

where is the initial momentum of the ping-poll ball

is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

is the final momentum of the ping-poll ball

is the final momentum of the bowling ball

We can re-arrange the equation as follows or

which means (1) so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

(2) therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

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heat excites molecules at their "lattice" sites. Enough to break the lattice bonds set the molecules free of each other and ... melt.

8 0
3 years ago
A 47 kg mass is moving across a horizontal surface at 8 m/s. What is the force required to bring the mass to a stop in 4.1 secon
crimeas [40]

Answer:

Force = -91.7 Newton

Explanation:

Given the following data;

Mass = 47 kg

Time = 4.1 seconds

Initial velocity = 8 m/s

Since the object comes to a stop, its final velocity would be equal to zero.

To find the force required to bring it to stop;

First of all, we would determine the acceleration of the object;

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

Substituting into the equation;

a = \frac{0 - 8}{4.1}

a = \frac{-8}{4.1}

Acceleration, a = -1.95 m/s²

Next, we would determine the force required to bring the object to stop;

Force = mass * acceleration

Force = 47 * -1.95

Force = -91.65 ≈ 91.7 Newton

6 0
3 years ago
"when you double the vehicle's weight, you will __________ the vehicle's stopping distance. "
Stolb23 [73]

With same braking power you will be stopping faster on the original weight therefore the answer to fill the blank is increase. The stopping distance will increase as there'll be higher energy to dissipate than lighter cars applied with the braking force similar with that of the lighter car. Also the skid and drag will add to the distance as well as the inertia of the moving heavier vehicle would be greater as well.

5 0
3 years ago
A tennis racquet hits a tennis ball. Why doesnt the racquet swing backwards when the ball hits it? (Shouldn't it swing back beca
sattari [20]

Answer:

You are putting more force on the ball no the racket so the racket wouldnt swing back but if you hit a racket on the ground it will swing backits just the way the force reaction works

Explanation:

^^^^^^^^^^^^^^^

4 0
3 years ago
A box at rest has the shape of a cube 3.5 m on a side. This box is loaded onto the flat floor of a spaceship and the spaceship t
guapka [62]

Answer:

V_o=25.725\ m^3

Explanation:

Given:

sides of the cube, a=3.5\ m

speed of the cube with respect to the observer, v=0.8c

Since the relative velocity of the object is relativistic, so there will be a length contraction according to the observer:

a_o=a\div\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }

where:

a_o= observed length of the side along the direction of velocity

a_o=3.5\div\frac{1}{\sqrt{1-\frac{(0.8c)^2}{c^2} } }

a_o=2.1\ m is the observed length of the cube edge only in the direction of the velocity due to relativistic effect of length contraction.

So the observed volume will be:

V_o=a\times a\times a_o

V_o=3.5\times 3.5\times 2.1

V_o=25.725\ m^3

5 0
3 years ago
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