PostLab 6_ Conservation of Momentum

This account cannot be access X Pf Sign On Course Home V engineering.pdf openvellum.ecollege.com/course.html?courseld=17215962&OpenVellumHMAC=d17f81cd781bbdbc830fe3efc4e25812#10001 Assignments Constants Syllabus Scores Part A A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring 15.0 .cm Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm .(Figure 1) еТext Find the magnitude of the block's velocity just after impact. Study Area ΑΣφ ? Document Sharing User Settings V = m/s Course Tools > Submit Request Answer Part B What was the initial speed of the bullet? ΑΣφ. ? v = m/s Submit Request Answer Figure 1 of 1 Provide Feedback Next > K15.0 cm- P Pearson Copyright © 2022 Pearson Education Inc. All rights reserved. | Terms of Use Privacy Policy Permissions | Contact Us |

USING THE UAM FORMULA, SOLVE THIS 1. A 6 kg object accelerates at 2.5 m/s^2 after a force is applied for 10s. Find the(a) Force applied.  (b) speed of the object at the end of 10 s,  (c) change in momentum of the block, and  (d) impulse produced by the force. 2. How much force is needed to stop a 10 kg object moving at 4 m/s in just 10 s? Explain your answer.

rlal BIUA 12 + 三 1 I will break you all into breakout rooms but each of you need to submit a copy Unit 3: Interactions and Conservation of Momentum 1. Students will show works for this problem to receive credits: A 25 kg wood box is sliding across a wood surface, experiencing a coefficient (u) of kinetic friction of 0.20. The box is experiencing an applied force of +63 Newtons. What is the force of friction acting on the box? (Ff=u x Fn), (Fn=-Fg), (Fg=mass x-9.81). P = mx v p= momentum m=mass V= velocity Directly proportional vs Inversely Proportional: When two variables are on different sides of the equal sign ( the equation), -if they are both on top of the fraction or both on the bottom of the fraction– directly proportional (Fc and m)( Fc and Vt) !!!

Learning Goal: To understand the relationship between force, impulse, and momentum. The effect of a net force EF acting on an object is related both to the force and to the total time the force acts on the object. The physical quantity impulse J is a measure of both these effects. For a constant net force, the impulse is given by J = FAt. The impulse is a vector pointing in the same direction as the force vector. The units of J are N. sor kg · m/s. Recall that when a net force acts on an object, the object will accelerate, causing a change in its velocity. Hence the chiect's momentum in- mail will also change. The Figure F (N) L 12,000 0.0007 1 of 1 t (s) In Parts A, B, C consider the following situation. In a baseball game the batter swings and gets a good solid hit. His swing applies a force of 12,000 N to the ball for a time of 0.70 × 10-³ s. Part A Assuming that this force is constant, what is the magnitude J of the impulse on the ball? Enter your answer numerically in newton…

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QUESTION 42 You are measuring the final momentum of two objects that stick together after colliding in an experiment. In several trials, you obtain the following values in kg·m/s: [20,24,32,17,15,26,24,27,40,36]. (b) In Excel, enter the data in a column and find the standard deviation and report the answer below and round up to the nearest whole number for example 13.99 would be 14. 14.32 would be 14.

p = mv J = Ft Ft = m Av Calculate the change in momentum of a 0.5 kg ball that strikes the floor at 15 m/s and bounces back up at 12 m/s. N

C Calculator X O * Dashboard | Khan Academy x A Week 6 Physics Quiz Impulse, x E Week 6 Ph LSdGd-dk1 DxxEx8sLZMSsGLCYS3uVGMsC StoTTKbm-mqCLvgw/viewform?hr_submission=DChkl403_3. Period 2-Physics 1. a) What is the momentum of a 5-kg bowling ball rolling at 4 m/s? b) If the bowling ball rolls into a pillow and stops in 1.5 s, calculate the average force on the pillow. c) What average Force does the pillow exert on the vectors required. 30 points ball. SKETCHES with

Every problem has a solution, Solve it! Instruction: Carefully read the word problem then identify the given, what is asked in the problem and provide solution. Write your answer with solution on a separate sheet of paper. 1. Four objects are situated along y-axis as follows; a 2.00-kg object is at + 3.00 meters, 5.00-kg object at 1.5 meters, a 4.5 – kg object at 3.0 meters and a 2.5 -kg object at 0.50 meters. Locate the center of mass of the system. 2. A billiard ball with a mass (m1) = 0.75-kg moves at a velocity vị of -2.0 m/s collides head-on with another billiard ball with a mass (m2) of =0.65 kg that is initially at rest (v2=0m/s) on the billiard table. No external forces act on the balls and collision is elastic. After collision, mi moves at the velocity of m2 velocity is equal to while the Note: To solve for vfi and vp we have these equations: where, Vf1 = mj-m2 : uy = final velocity/ velocity after collision Vi1 m2 2m, mị + m2 Vi = initial velocity/velocity before collision…

1. Which of the following statements is/are TRUE? a. The coefficient of restitution can take up a value greater than 1. b. The center of mass can be located outside the object, especially those with curved structures. c. The initial momentum of a moving body is directly proportional to the force applied and to the duration of its application. d. The final velocities of two colliding objects can be calculated if the coefficient of restitution is given along with the objects' initial velocities.

6.3 PQ1 Newton's second law can be stated that the force is equal to the rate of change of momentum, F = d(mv)/dt. Tak- ing the derivative by parts yields F = m(dv) / (dt) + v(dm) / (dt). This does not correspond to F = ma. What is the source of the discrepancy? %3D

1) A 100 g cart initially moving at +0.5 m/s collides elastically from a stationary 180 g cart. a) Using the equation in the theory, calculate the final velocity of the 100 g cart. b) Calculate the change in the momentum of the 100 g cart. c) Since this is an elastic collision, total momentum must be conserved. Using your answer for part b, calculate the final velocity of the 150 g cart.

Answer the following questions: 1. A 32 kg child sits at one end of a 3.8 m long seesawWhere should her 72 kg father sit so the center of mass will be at the center of the seesaw? 2. A firework explodes into three pieces as shown in the diagram below. Now that you have set up the equation for the x and y components of the final momentum of the firework, let's do some physics! Given: ∙∙ The initial firework had a mass of 9.0 kg and was launched at vi = 50 m/s vertically straight up. ∙∙ The firework explodes into three pieces of equal mass 3.0 kg. The angles θ1 = θ3 = 40 ∘. ∙∙ Speed v1 = 100 m/s.   What is the speed v2?   Note: the angles shown for mass m1 and mass m3 are relative to the horizontal x-axis. Mass m2 is moving straight up (y-axis).

(a) What is the mass of a large ship that has a momentum of 1.40 x 109 kg-m/s, when the ship is moving at a speed of 38.0 km/h? 1.327 X kg (b) Compare the ship's momentum to the momentum of a 1400 kg artillery shell fired at a speed of 1300 m/s. P ship 769.23 Pshell Additional Materials MReading

a) What impulse occurs when an average force of 10 N is exerted on a cart for 2.5 s? b. What change in momentum does the cart undergo? c. If the mass of the cart is 2 kg and the cart is initially at rest , calculate its final speed.

Good day. Can you please inspect my solution. Is it wrong? I used the summation of forces=momentum change.

34. In trial 1 of an experiment, a cart moves with speed vo on a frictionless, horizontal track and collides elastically with another cart that is initially at rest. In trial 2, the setup is identical except that the carts stick together during the collision. How does the speed of the two-cart system's center of mass change, if at all, during the collision in each trial? Trial 1 Trial 2 (A) Does not change (B) Does not change C) Decreases (Đ) Decreases Does not change Decreases Does not change teg Decreases (A) ganga oold sd

9:07 fing n O O X * ll 71% Pearson https://openvellum.ecollege.com/cou. Course Home (MP11-19: Momentum and Impulse 221-16-3 3 of 4 I Review | Constants Part A Use the impulse-momentum theorem to find how long a falling object takes to increase its speed from 5.60 m/s to 11.0 m/s. ΑΣΦ S Submit Request Answer Provide Feedback Next > P Pearson

Directions: Select whether the statement is true of false. 1. As the ball rises, it decreases its vertical velocity.2. The projectile experiences a constant vertical acceleration due to the pull of gravity as it rises and falls.3. Collision conserved momentum.4. The total time (t’) of flight is simply double the time it takes for the projectile to rise.5. Perfectly inelastic collision is described by maximum lost in kinetic energy.6. Elastic collision happens when 2 objects stick together after collision.7. Elastic collision conserved kinetic energy.8. The time it takes to rise to its highest point is longer than the time it takes to fall. 9. Kinetic energy is conserved in all kinds of collision.10. The horizontal distance known as the range, R of the projectile is the ratio of the horizontal velocity and the total time of flight.

Le Style APA Bibliography Update Table L Update Index LUpdate Table Mark Citation Insert Mark Caption C Cross-reference Entry ations & Bibliography Captions Index Table of Authorities 3. a) What is the difference between inertia and momentum? b) A bullet of mass 40.0 g is fired from a gun with a nozzle velocity of 25.0 ms1. Calculate the recoil velocity of the gun whose mass is 5.0 kg

4. You need to show your work in detail to get any credit. Before collision: The red car (on the left) has the velocity of 10.0 m/s and mass of 4.8 kg. The blue car (on the right) has the velocity of -5.6 m/s and mass of 4.4 kg. After elastic collision: The blue car moves with the velocity of 8.8 m/s. (a) What is the initial momentum of the red car? (b) What is the initial momentum of the blue car? (c) What is the total momentum of the system of blue and red car before collision? (d) What is the final momentum of the blue car? (e) What is the impulse on the blue car (in kg.m/s)? (f) Is the system consisting of the blue and red car an isolated system? (g) What is the impulse on the red car (in kg.m/s)? (h) What is the velocity of the red car after the collision? (i) Draw annotated p-t and F-t graphs for both carts at all stages of collision. Assume that collision occurred for 0.5 seconds. Before Р Ft During +X After 24 t 4 t

A 50 kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s. a) What is the initial momentum of the mass?b) What is the final momentum of the mass?c) What was the force acting on the mass?d) What was the impulse acting on the mass?

A rigid car body of mass 2346.5x10-2 kg which moving velocity its velocity changed from 14 m/sec to 31 m/sec in the same direction with time in 66 seconds.  Calculate the (i) Change in its momentum in kgm/sec and (ii) Force which is responsible for this momentum variation in newton.  Change in momentum in kgm/sec   Force of the body in kN

Activity 4: Solve the following problems. A 2600 kg SUV traveling west at 15 m/s collides head on with 1300 kg sedan traveling east at 20 m/s. Determine the speed and direction of the two cars immediately after impact if they lock together? 1. Before Collision After Collision Object A Mass = Mass = Velocity =D Total Momentum%3Dmass x velocity Velocity = Total Momentum3Dmass velocity P3mv P=mv Object B Mass = Mass = Velocity = x Total Momentum3Dmass x velocity Velocity = Total Momentum3mass velocity P=mv P=mv Total Total Momentum Total Momentum (MaxVb)+(MpxV)= The Deşired Equation for the Problem/Solution

2 The shown square plate is made from two different materials vith densities of pi and p2 respectively with negligible thickness. If pr: P2 1:3 find the center of mass. За Pi P2 За

The brakes on a 15,680 N car exert a stopping force of 640. N. The car’s velocity changes from 20.0 m/s to 0 m/s. a) What is the car’s mass? b) What was its initial momentum? c) What was the change in momentum for the car? d) How long does it take the braking force to bring the car to rest?

Maste session.masteringphysics.com/myct/itemView?assignmentProblemlD3D136303042&offset%3Dnext Apps (1) .. * YouTube to MP3 Co... phone cover for mo... Online Services Fo... AGARWAL-PHY-166 CHW6 Item 12 A 144-g baseball moving 28 m/s strikes a stationary 5.25-kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bounces straight back, and the brick moves forward at 1.38 m/s. ACbefore value UNIIS Submit Request Answer Part C Determine the total kinetic energy after the collision. Express your answer to two significant figures and include the appropriate um НА KEafter = Value Units %3D Submit Request Answer Provide Feedback