A 1.95-kg particle has a velocity (1.95 î – 3.07 ĵ) m/s, and a 2.91-kg particle has a velocity (1.06 î + 5.99 ĵ) m/s.(a) Find the velocity of the center of mass.i) m/s(b) Find the total momentum of the system.|î) kg · m/s

Name: A 1.95-kg particle has a velocity (1.95 î – 3.07 ĵ) m/s, and a 2.91-k
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: A 1.95-kg particle has a velocity (1.95 î – 3.07 ĵ) m/s, and a 2.91-kg particle has a velocity (1.06 î + 5.99 ĵ) m/s.(a) Find the velocity of the center of mass.i) m/s(b) Find the total momentum of the system.|î) kg · m/s

Answered: Image /qna-images/answer/d4eac007-094b-466d-a5ca-64dc2273b17d.jpg

Science

Physics

A 1.95-kg particle has a velocity (1.95 î - 3.07 j) m/s, and a 2.91-kg particle has a velocity (1.06 î + 5.99 j) m/s. (a) Find the velocity of the center of mass. m/s (b) Find the total momentum of the system. i) kg · m/s

A 1.95-kg particle has a velocity (1.95 î - 3.07 j) m/s, and a 2.91-kg particle has a velocity (1.06 î + 5.99 j) m/s. (a) Find the velocity of the center of mass. m/s (b) Find the total momentum of the system. i) kg · m/s

Classical Dynamics of Particles and Systems

5th Edition

ISBN:9780534408961

Author:Stephen T. Thornton, Jerry B. Marion

Publisher:Stephen T. Thornton, Jerry B. Marion

Chapter9: Dynamics Of A System Of Particles

Section: Chapter Questions

Problem 9.60P

See similar textbooks

Related questions

Q: Starting at t = 0, a horizontal net force F=(0.290N/s)t(i)+(−0.450N/s^2)t^2(j) is applied to a box…

Q: In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table.…

A: The 3.5 g bullet is moving horizontally towards two blocks kept at rest. The bullet first penetrates…

Q: 20 m/s to the east relative to Earth. It collides head-on with

A: To solve for the final velocities of the balls, we can use the conservation of momentum and the…

Q: A 5.50 g bullet moving at 980 m/s strikes a 510 g wooden block at rest on a frictionless surface.…

Q: An object with mass m1 = 3.00 kg is moving along the positive x axis with a speed v1i = 2 m/s…

Q: 8. An object (A) of mass mA = 30.5 kg is moving in a direction that makes angle of 60° south of west…

Q: A 35.2 kg object takes 41.9 s to stop from an initial speed of 5.3 m/s. What is the magnitude of the…

A: Given data- mass of object (m) = 35.2 Kg initial speed of object ( u ) = 5.3…

Q: A particle of mass m, = 1.50 kg and initial velocity U1, i = (9.30 i + 8.00 j) m/s collides with the…

Q: A 0.70 kg ball moving horizontally at 5.0 m/s strikes a vertical wall and rebounds with speed 2.0…

A: given : Mass m =0.75 kg v1= 5 m/s v2 = -2 m/s

Q: Current Attempt in Progress A vessel at rest at the origin of an xy coordinate system explodes into…

Q: The figure below gives an overhead view of the path taken by a 0.165 kg cue ball as it bounces from…

A: Given data: Mass of ball, m=0.165 kg Initial speed of ball, u=1.70 m/s Angle, θ1=300

Q: A particle with mass ma = 3.00 kg is located at ra = (2.50 i+ 3.50 j) m, and a second particle of…

Q: A sliotar of mass 115 g moves horizontally with an initial velocity of 38.0 m/s as it approaches a…

A: Mass = 0.115 kg Initial Velocity (u) = 38 m/s Final velocity (v) = -58 m/s Time (t) = 12.1 ms =…

Q: In the figure below, m1 3kg, m2= 4kg, and m3 = 8kg, and the scales on the axes are set by xs = 2m…

A: In given figure m1=3 kg at x1=0 and y1=0m2=4 kg at x1=2 m and y1=1 mm3=8 kg at x1=1 m and y1=2 m x…

Q: A bullet of mass 27.0 grams is shot horizontally into a fong rod with a wooden block of mass 2.00 kg…

A: Option (c) is correct that is 53 degrees.

Topic Video

Question

A 1.95-kg particle has a velocity (1.95 î – 3.07 ĵ) m/s, and a 2.91-kg particle has a velocity (1.06 î + 5.99 ĵ) m/s.
(a) Find the velocity of the center of mass.
i) m/s
(b) Find the total momentum of the system.
|î) kg · m/s

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

The vector position of a 3.50-g particle moving in the xy plane varies in time according to r1=(3i+3j)t+2jt2, where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.50 g particle varies as r2=3i2it26jt. At t = 2.50 s, determine (a) the vector position of the center of mass of the system, (b) the linear momentum of the system, (c) the velocity of the center of mass, (d) the acceleration of the center of mass, and (e) the net force exerted on the two-particle system.
A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass ml = 48.0 kg travels in the positive x-direction at 12.0 m/s, and a second piece of mass m2 = 62.0 kg travels in the xy-plane at an angle of 105 at 15.0 m/s. The third piece has mass m3 = 112 kg. (a) Sketch a diagram of the situation, labeling the different masses and their velocities, (b) Write the general expression for conservation of momentum in the x- and y-directions in terms of m1, m2, m3, v1, v2 and v3 and the sines and cosines of the angles, taking to be the unknown angle, (c) Calculate the final x-components of the momenta of m1 and m2. (d) Calculate the final y-components of the momenta of m1 and m2. (e) Substitute the known momentum components into the general equations of momentum for the x- and y-directions, along with the known mass m3. (f) Solve the two momentum equations for v3 cos and v3 sin , respectively, and use the identity cos2 + sin2 = 1 to obtain v3. (g) Divide the equation for v3 sin by that for v3 cos to obtain tan , then obtain the angle by taking the inverse tangent of both sides, (h) In general, would three such pieces necessarily have to move in the same plane? Why?
A 2-kg object moving to the right with a speed of 4 m/s makes a head-on, elastic collision with a 1-kg object that is initially at rest. The velocity of the 1-kg object after the collision is (a) greater than 4 m/s, (b) less than 4 m/s, (c) equal to 4 m/s, (d) zero, or (e) impossible to say based on the information provided.
A 2.00-kg particle has a velocity (2.00i3.00j)m/s, and a 3.00-kg particle has a velocity (1.00i+6.00j)m/s. Find (a) the velocity of the center of mass and (b) the total momentum of the system.
The magnitude of the net force exerted in the x direction on a 2.50-kg particle varies in time as shown in Figure P9.10 (page 244). Find (a) the impulse of the force over the 5.00-s time interval, (b) the final velocity the particle attains if it is originally at rest, (c) its final velocity if its original velocity is 2.00im/s, and (d) the average force exerted on the particle for the time interval between 0 and 5.00 s. Figure P9.10
A 2.00-kg particle has a velocity (2.00. 3.00) m/s, and a 3.00-kg particle has a velocity (1.00 + 6.00) m/s. Had (a) the velocity of the center of mass and (b) the total momentum of the system.
A car of mass 1 560 kg traveling east and a truck of equal mass traveling north collide and become entangled, moving as a unit at 15.0 m/s and 60.0 north of east. Find the speed of (a) the car, and (b) the truck prior to the collision. (See Section 6.4.)
A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially unstretchcd and with force constant k = 2.00 104 N/m, as shown in Figure P8.60. The cannon fires a 200-kg projectile at a velocity of 125 m/s directed 45.0 above the horizontal. (a) Assuming that the mass of the cannon and its carriage is 5 000 kg, find the recoil speed of the cannon. (b) Determine the maximum extension of the spring. (c) Find the maximum force the spring exerts on the carriage. (d) Consider the system consisting of the cannon, carriage, and projectile. Is the momentum of this system conserved during the firing? Why or why not?
A tennis ball of mass 57.0 g is held just above a basketball of mass 590 g. With their centers vertically aligned, both balls are released from rest at the same time, falling through a distance of 1.20 m, as shown in Figure P6.45. (a) Find the magnitude of the basketballs velocity the instant before the basketball reaches the ground. (b) Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball so that it collides with the tennis ball just above it. To what height does the tennis ball rebound? Figure P6.45
What exhaust speed is required to accelerate a rocket in deep space from 800 m/s to 1000 m/s in 5.0 s if the total rocket mass is 1200 kg and the rocket only has 50 kg of fuel left?
A model rocket engine has an average thrust of 5.26 N. It has an initial mass of 25.5 g, which includes fuel mass of 12.7 g. The duration of its burn is 1.90 s. (a) What is the average exhaust speed of the engine? (b) This engine is placed in a rocket body of mass 53.5 g. What is the final velocity of the rocket if it were to be fired from rest in outer space by an astronaut on a spacewalk? Assume the fuel burns at a constant rate.
A 2.09-kg particle has a velocity (1.90 î – 2.90 j) m/s, and a 2.95-kg particle has a velocity (0.92 î + 6.01 j) m/s. (a) Find the velocity of the center of mass. |î + i) m/s (b) Find the total momentum of the system. i) kg · m/s