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algol [13]
3 years ago
14

Consider two balls on a horizontal, frictionless surface. Ball A (with mass ma) is moving toward ball B (with mass mb = 1 kg). B

all B is initially at rest.a. (5 pts) If their collision is perfectly elastic, what should be the mass of ball A such that it stops upon collision if ball A's initial velocity is +3.1416 m/s?b. (5 pts) Using conditions in question (a), what is the speed of ball B after collision? Is it moving in the initial direction of ball A or opposite it?c. (5 pts) Consider this 2nd scenario: If the collision is perfectly inelastic, what is the speed of the balls after collision? [Use the mass that you calculated from question (a) for ball A.]d. (5 pts) Now, consider a 3rd scenario: Ball A (with arbitrary speed va) collides with a stationary ball B (mass is mb = 1 kg). After collision, you observed that ball A and ball B are now moving at the same speed va but toward opposite directions: Ball A is heading opposite its original direction, while B is moving along the initial direction of A. What is the mass of ball A?
Physics
1 answer:
Sonbull [250]3 years ago
4 0

Answer:

Explanation:

When two ball of identical mass collides perfectly elastically , there is exchange of velocity between the two balls .

Here ball be was at rest initially . After collision ball A comes to rest , so there is complete exchange of velocity . Hence ball A must have same mass as that of B . mass of ball A = 1 kg .

b ) Due to complete exchange of velocity , velocity of ball A will be picked up by ball B . Hence velocity of ball B = 3.1416 m /s . Yes it will be moving in the direction of ball A .

c )

In case of perfectly inelastic collision they will become  single mass

total mass = 2 kg

applying conservation of momentum law

their common velocity after collision = 1 x 3.1416 / 2 = 1.57 m /s

d )

Applying conservation of momentum law

initial momentum = Ma x va

they move in opposite direction after collision

their total momentum after collision

1 x va - Ma va

applying law of conservation of momentum

1 x va - Ma va  = Ma va

va = 2Ma va

Ma = .5 kg .

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The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m
algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

E=PE+KE=const.

The potential energy is given by

PE=\frac{1}{2}kx^2

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

E=KE_{max}=\frac{1}{2}mv_{max}^2 (1)

where

m is the mass of the block

v_{max}=1.25 m/s is the maximum speed

We also know that the total energy is

E=0.18 J

Re-arranging eq.(1), we can find the mass:

m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg

b)

The maximum speed in a spring-mass system is also given by

v_{max} =\sqrt{\frac{k}{m}} A

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

v_{max}=1.25 m/s is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m

c)

The frequency in a spring-mass system is given by

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz

d)

We can solve this part by using the law of conservation of energy; in fact, we have

E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s

#LearnwithBrainly

3 0
3 years ago
Which best describes the type of lens that would help someone with one of these eye conditions? Concave lenses would help someon
s2008m [1.1K]

Answer:

The answer is "Option a".

Explanation:

The concave lenses are there for narrow-sighted people, convex for the far people. In eyewear, curved lenses are used to rectify visibility. Since in nearsighted individuals the width between the eye lens as well as the retina is larger than should be, these readers can't clearly distinguish distant things.

7 0
3 years ago
Read 2 more answers
a set of water travels at 10 m/s, and 5.0 waves pass you in 2.0s. what is the wavelength of the waves?
Alecsey [184]
Well, I think the answer would be 4 meters because the speed is in meters per second and the frequency is 5 waves every 2 seconds. 
Wavelength = Speed/Frequency
I think the frequency is 2.5 waves a second but that's not something I'm 100% sure about.
I hope this helps though.
3 0
4 years ago
Explain the difference between how intrusive and extrusive igneous rock form and how those formation processes affect the minera
rusak2 [61]

Extrusive molten rocks are formed from lava while the intrusive molten rocks are formed from magma.

<u>Explanation:</u>

Extrusive molten rocks originate from lava, framing at the outside of the Earth and cooling rapidly, which means they structure little precious stones. Intrusive volcanic rocks originate from magma, shaping profound underground and taking more time to cool, which means they structure bigger crystals.

Intrusive molten rocks are framed when magma cools gradually underneath the surface, while extrusive rocks are shaped when magma cools quickly at the surface.

6 0
4 years ago
An object moves in uniform circular motion at 25 m/s and takes 1.0 second to go a quarter circle. Calculate the centripetal acce
tekilochka [14]

Given:

Object in circular motion 25 m/s

1 second to go quarter circle

Required:

Centripetal acceleration:

Solution:

Acceleration = v2/r

Where v is the velocity and r is the radian

Substituting the values into the equation,

Acceleration = v2/r = (25 m/s)2/(4*pi/180) = 8952.47 m2/s2

4 0
4 years ago
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