a lump of putty and a rubber ball have equal mass. both are thrown with equal speed against a wall. the putty sticks to the wall. the ball bounces back at nearly the same speed with which it hit the wall. which object experiences the greater momentum change?

Answers

Answer 1
Answer: The bouncy ball experiences the greater momentum change.

To understand why, you need to remember that momentum is actually
a vector quantity ... it has a size AND it has a direction too.

The putty and the ball have the same mass, and you throw them
with the same speed.  So, on the way from your hand to the wall,
they both have the same momentum.
Call it " M in the direction toward the wall ".

After they both hit the wall:

-- The putty has zero momentum.
    Its momentum changed by an amount of  M .

-- The ball has momentum of " M in the direction away from the wall ".
    Its momentum changed by an amount of  2M .
Answer 2
Answer:

Rubber ball will experience the greater momentum change than the lump of putty

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Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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Given:

mass of a lump of putty = m

mass of rubber ball = m

initial speed of putty = v

final speed of putty = 0

initial speed of rubber ball = v

final speed of rubber ball = -v

Asked:

the change in momentum = Δp = ?

Solution:

Lump of Putty :

\Delta p_(putty) = mv_f' - mv_i

\Delta p_(putty) = m(0) - mv

\Delta p_(putty) = - mv

\texttt{ }

Rubber Ball:

\Delta p_(ball) = mv_f' - mv_i

\Delta p_(ball) = m(-v) - mv

\Delta p_(ball) = - 2mv

\texttt{ }

\Delta p_(ball) : \Delta p_(putty) = - 2mv : - mv

\Delta p_(ball) : \Delta p_(putty) = 2 : 1

\Delta p_(ball) = 2\Delta p_(putty)

\Delta p_(ball) > \Delta p_(putty)

From calculation above , we can conclude that rubber ball will experience the greater momentum change than the lump of putty

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Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant


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The Earth has a radius of 6,400 kilometers. A satellite orbits the Earth at a distance of 12,800 kilometers from the center of the Earth. If the weight of the satellite on Earth is 100 kilonewtons, the gravitational force on the satellite in orbit is?It would be great if you could add a few words of explanation.

Answers

By using a universal gravitational force, we can get that F_(g) = (G. m_(1) . m_(2))/( r^(2) )

For note : r is a distance from center of the earth to the object.

Then, if r_(1) is 6.400 km, and if r_(2) is 12.800 km,
we can say that r_(2) is 2r_(1)

In first equation we can say that :
F_(g) = (G.m_(1) m_(2) )/( r_(1)^(2) ) = 100.000 N

then in second equation we can say that :
F_(g) = (G.m_(1) m_(2) )/( r_(2)^(2) )
F_(g) = (G.m_(1) m_(2) )/( (2r_(1))^(2) ) (because r_(2) is 2r_(1) )
F_(g) = (G.m_(1) m_(2) )/( 4r_(1)^(2) )
we can say that : F_(g) = (1)/(4) (G.m_(1) m_(2) )/( r_(1)^(2) )
so, by plugging first equation into second equation, we can get
F_(g) = (1)/(4) (G.m_(1) m_(2) )/( r_(1)^(2) ) = (1)/(4) . 100.000 N = 25.000 N

You observe a distant galaxy. you find that a spectral line normally found in the visible part of the spectrum is shifted toward the infrared. what do you conclude?

Answers

The galaxy is moving away from you.

An object has a mass of 120 kg on the moon. What is the force of gravity acting on the object on the moon? A. 12.24 N
B. 20 N
C. 1,176 N
D. 196 N User: A man standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down the man continues moving forward at the original speed. This is an example of the effect of
A. velocity.
B. gravity.
C. weight.
D. inertia.

Answers


Well first of all, if the object's mass is 120 kg, then its mass is 120 kg. 
It makes no difference where it happens to be located at the moment. 
Its mass is 120 kg. Period.  It doesn't change.

Weight = (mass) x (acceleration of gravity) .

Acceleration of gravity on the moon is 1.622 m/s² (rounded) .

Weight of the object = (120 kg) x (1.622 m/s²) = 194.6 Newtons.
                                                                   (about 43.8 pounds)

The nearest choice is  D). 196 N.

===========================================

When the bus starts moving forward, the man remains still. 
He leans back, and soon catches up.

When the bus is moving but begins to slow down, the man
continues moving forward at the same speed.  He leans forward,
until the bus stops, and then he eventually stops.

The simplest way to explain this (although it doesn't tell you much)
is to invoke 'inertia'.  'D' is the choice of the choices.

For first question, Gravity is approximately up to 1/6 of Earth's gravity. So object's weight 1200/6=200 N. So answer shall be D.

For second question, intertia is the thing that described in the question. Answer is D.

An 8.10 kilogram box of mass is kept at a height of 0.99 meters above ground level. What is the potential energy of the box? (given g = 9.8 meters/second2). A. 68 joules. B. 79 joules. C. 81 joules. D. 99 joules. E. 1.0 × 102 joules

Answers

The answer is B. 79 joules.

To calculate this we will use the formula for potential energy (PE):
PE = m * g * h
where m is mass of an object, g is gravitational acceleration, h is a height at which is the object.

It s given:
m = 8.10 kg
g = 9.8 m/s²
h = 0.99 m

After replacing them in the formula for potential energy:
PE = 8.10 * 9.8 * 0.99 = 78.6 J ≈ 79 J

The answer would be 79J because when you multiply 8.1*0.99*9.8 you get 7.89, which rounds to 7.9.

Answer choice (B).

. A ray of monochromatic light (f = 5.09 × 1014hertz) in air is incident at an angle of 30.°
on a boundary with corn oil. What is the angle of refraction, to the nearest degree, for this
light ray in the corn oil?

Answers

The angle of refraction of light ray in the corn oil is 20°.

What is angle of Refraction?

Angle of refraction is the angle the refracted light ray makes with the normal to the oil-air interface.

Given data:

Angle of incidence = 30°

Analysis:

The refractive index of oil is 1.47.

Using the formula for refractive index(n) = (sini)/(sinr)

1.47 = (sin 30)/(sin r)

1.47(sin r) = sin 30

sin r = (sin 30)/(1.47)

sin r = 0.3401

r = arc sin(0.3401)

r = 19.88° which is approximately 20°

Learn more about Refraction : brainly.com/question/10729741

#SPJ2

Answer:

its 20 degrese

Explanation:

A ski jumper competing for an Olympic gold medal wants to jump a horizontal distance of 135 meters. Thetakeoff point of the ski jump is at a height of 25 meters. With what horizontal speed must he leave the jump?
a. What do you know?
b. What do you need to solve for?
c. What equation(s) will you use?
d. What is the solution to this problem?

Answers

A. We know that the height (h) = 25 m, and the distance (x) = 135 m.

B. We need to find the time flight first (t), then we can find easily the horizontal velocity ( v_(x). )

C. Because this is a projectile motions, so horizontal velocity is constant, but the vertical  velocity is change because of the gravitational acceleration.
So we can use : h = (1)/(2) g t^(2),

Note : Time that will be exist is a time from a land to the takeoff point, that can be called as the time flight.
Then we use, x = v_(x) t, to find the horizontal velocity.

D. We know that :
     h = (1)/(2) g t^(2), so by insert the numbers, and assune that the (g = 10 m/s^(2)), we get that ( t = √(5). )
     After that by using  x = v_(x) t, we can get that v_(x) = 27 √(5)