= 4.05 kq is released from . It makes a head-on elastic collision at B with a block ofConsider a frictionless track as shown in the figure below. A block of mass m13.0 kg that is initially at rest. Calculate the maximum height to which m, rises after the collisionmass m2mтy5.00 mт2B

Question
Asked Oct 29, 2019
55 views
= 4.05 kq is released from . It makes a head-on elastic collision at B with a block of
Consider a frictionless track as shown in the figure below. A block of mass m
13.0 kg that is initially at rest. Calculate the maximum height to which m, rises after the collision
mass m2
m
тy
5.00 m
т2
B
help_outline

Image Transcriptionclose

= 4.05 kq is released from . It makes a head-on elastic collision at B with a block of Consider a frictionless track as shown in the figure below. A block of mass m 13.0 kg that is initially at rest. Calculate the maximum height to which m, rises after the collision mass m2 m тy 5.00 m т2 B

fullscreen
check_circle

Expert Answer

Step 1

Apply conservation of energy before collision. After release of the block to before the collision in order to obtain the velocity of the block of mass m1 before collision.

help_outline

Image Transcriptionclose

КЕ, + РЕ, 3 КЕ, + РE, 1 0+ тgh%3D- m 2 0 "л2gh /2(9.81m/s2 )(5.00 m) 9.05 m/s

fullscreen
Step 2

The direction of velocity of the block will along horizontal before collision.

Apply the conservation of momentum just before the collision and after the collision.

help_outline

Image Transcriptionclose

= mv,1 +m2v?,2 MVFIMV (4.05 kg) (9.05 m/s2) + (13.0 kg) (0 m/s) = (4.05 kg)v',, +(13.0 kg) v',2 = 36.65 m/s 4.05v13.0 1 (36.65 m/s - 4.05v1 13

fullscreen
Step 3

Apply conservation of kinetic energy...

help_outline

Image Transcriptionclose

1 - тv" 2 1 - ту 2 2 my+mymm ту туй + т. v + т,v, — (4.05 kg) (9.05 m/s)(13.0 kg) (0 m/s)' (4.05 kg)v% +(13.0 kg)v% = 331.70 m2/s 4.05v13.0

fullscreen

Want to see the full answer?

See Solution

Check out a sample Q&A here.

Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in

Science

Physics

Work,Power and Energy

Related Physics Q&A

Find answers to questions asked by student like you
Show more Q&A
add
question_answer

Q: Find the period (T), frequency (f), and angular frequency (ω) of the oscillation.T =  _____ sf =  __...

A: From the plot of the question .

question_answer

Q: You are very particular about the proper temperature needed to make tea. The proper temperature for ...

A: The temperature of the boiling water is 100 degree Celsius.Convert the temperature of favorite tea t...

question_answer

Q: A 51.0 g drink is left on a turntable causing it to turn with an angular speed of 7.10 rad/s. If the...

A: Given:Mass of drink = 51.0 gAngular speed = 7.10 rad/sCoefficient of static friction, μs = 0.830Coef...

question_answer

Q: How do I solve this for "maximum amplitude of oscillation such that the top block will not slip on t...

A: Figure shows A block with mass M rests on a frictionless surface and is connected to a horizontal sp...

question_answer

Q: A flashlight bulb is rated 2.0 V, 0.23 A. How much power is required to operate the light bulb? Roun...

A: Power is,

question_answer

Q: A dragster starts from rest and accelerates down a track. Each tire has a radius of 0.320 m and roll...

A: a)Linear speed is,

question_answer

Q: Let's assume that two objects with masses mA and mB are mvoing with velocities vA and vB along the x...

A: Let va, vb are the initial velocities of masses a and b, and va’, vb’ are the final velocities of ma...

question_answer

Q: (b) What torque must be applied to this object to give it an angular acceleration of 1.30 rad/s2 if ...

A: Torque is,

question_answer

Q: An object of mass m1 = 3.50 kg placed on a frictionless, horizontal table is connected to a string t...

A: Given values: m1 = 3.50 kgm2 = 6.60 kg