1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
arsen [322]
3 years ago
13

A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i

s true?
The magnitude of the change of the momentum of the 5-kg ball is equal to the magnitude of the change of momentum of the 10-kg ball.
Both balls lose all their momentum since the collision is inelastic.
The magnitude of the change of velocity the 5-kg ball experiences is greater than that of the 10-kg ball.
The magnitude of the change of velocity the 5-kg ball experiences is equal to that of the 10-kg ball.
The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.
Physics
1 answer:
NARA [144]3 years ago
4 0

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = \frac{1}{2}mu^2

Final kinetic energy =\frac{1}{2}mv^2

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

You might be interested in
A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the
Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, \Delta t=0.003\ s

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N

So, the average force on ball by the golf club is 340 N.

4 0
4 years ago
Running up a hill instead of a flat surface will increase which principle of fitness?
musickatia [10]

Answer:

a

Explanation:

5 0
3 years ago
Read 2 more answers
A block of ice of mass 4.30 kg is placed against a horizontal spring that has a force constant k = 250 N/m and is compressed a d
OleMash [197]

Answer:

W = 0.060 J

v_2 = 0.18 m/s

Explanation:

solution:

for the spring:

W = 1/2*k*x_1^2 - 1/2*k*x_2^2

x_1 = -0.025 m and x_2 = 0

W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2

W = 0.060 J

the work-energy theorem,

W_tot = K_2 - K_1 = ΔK

with K = 1/2*m*v^2

v_2 = √2*W/m

v_2 = 0.18 m/s

8 0
3 years ago
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend
Viefleur [7K]

a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how

long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175

s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

is then twice this, or 0.08s

5 0
3 years ago
Read 2 more answers
Select all that apply. Which of the following do not illustrate a complete circuit? A B C D
ipn [44]
There is not enough information here.
7 0
3 years ago
Read 2 more answers
Other questions:
  • Jeff's body contains about 5.12 L5.12 L of blood that has a density of 1060 kg/m3.1060 kg/m3. Approximately 45.0%45.0% (by mass)
    10·1 answer
  • Fluids can flow and change shape to fit their containers. Which two states are<br> fluids?
    10·1 answer
  • _____ is the preceived frequency of a sound wave.​
    15·1 answer
  • What is the circumference of the moon?
    6·2 answers
  • A wave on a string is described by D(x,t)= (2.00cm)sin[(12.57rad/m)x−(638rad/s)t], where x is in m and t is in s. The linear den
    9·1 answer
  • PLEASE HELP ME
    5·1 answer
  • Physical Science - 02.05 - Question #3
    12·1 answer
  • According to Newton’s first law of motion what will an object in motion do when no external force acts on it ?
    14·1 answer
  • How many SF does number 0.00403 have?
    8·1 answer
  • A food processor draws 8.47 A of current when connected to a potential difference of 110 V.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!