(a)
The total kinetic energy of the two blocks.
(a)
Answer to Problem 42P
The total kinetic energy of the two blocks is
Explanation of Solution
Given:
The mass of block 1 is
The velocity of the block 1 in positive x direction is
The mass of block 2 is
The velocity of block 2 in positive x direction is
Formula used:
The expression for kinetic energy is given by,
Here,
Calculation:
The total Kinetic energy of the two blocks can be calculated as,
Further solving the above equation,
Conclusion:
Therefore, the total kinetic energy of the two blocks is
(b)
The velocity of centre of mass of the two block system.
(b)
Answer to Problem 42P
The velocity of centre of mass of the two block system is
Explanation of Solution
Given:
The mass of block 1 is
The velocity of the block 1 in positive x direction is
The mass of block 2 is
The velocity of block 2 in positive x direction is
Formula used:
The expression for momentum is given by,
Calculation:
Since, the momentum of the system remains conserved and can be calculated as,
Conclusion:
Therefore, the velocity of centre of mass of the two block system is
(c)
The velocity of each block relative to the centre of mass.
(c)
Answer to Problem 42P
The velocity of the block 1 relative to the centre of mass is
Explanation of Solution
Given:
The velocity of the block 1 in positive x direction is
The velocity of block 2 in positive x direction
Formula used:
The expression for the velocity with respect to centre of mass is given as,
Here,
Calculation:
The velocity of block 1 with respect to the centre of mass can be calculated as,
The velocity of block 2 with respect to the centre of mass can be calculated as,
Conclusion:
Therefore, The velocity of the block 1 relative to the centre of mass is
(d)
The kinetic energy of the blocks relative to the centre of mass.
(d)
Answer to Problem 42P
The kinetic energy of the blocks relative to the centre of mass is
Explanation of Solution
Given:
The velocity of block 1 in positive x direction is
The velocity of block 2 in positive x direction is
Formula used:
The expression for kinetic energy is given by,
Calculation:
The total Kinetic energy of the two blocks can be calculated as,
Further solving the above equation,
Conclusion:
Therefore, the kinetic energy of the blocks relative to the centre of mass is
(e)
The proof that the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.
(e)
Answer to Problem 42P
The kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.
Explanation of Solution
Given:
The mass of block 1 is
The velocity of the block 1 in positive x direction is
The mass of block 2 is
The velocity of block 2 in negative x direction
Formula used:
The expression for kinetic energy is given by,
Calculation:
The total Kinetic energy of the two blocks can be calculated as,
The above result is equal to
Conclusion:
Therefore, the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.
Want to see more full solutions like this?
Chapter 8 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
- In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling in the same direction. The initial speed of the leading bumper car is 5.60 m/s and that of the trailing car is 6.00 m/s. Assuming that the mass of the drivers is much, much less than that of the bumper cars, what are their final speeds?arrow_forwardConfirm that the results of the example Example 8.7 do conserve momentum in both the x- and y -directions.arrow_forwardA 2.00-g particle moving at 8.00 m/s makes a perfectly elastic head-on collision with a resting 1.00-g object. (a) Find the speed of each particle after the collision. (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10.0 g. (c) Find the final kinetic energy of the incident 2.00-g particle in the situations described in parts (a) and (b). In which case does the incident particle lose more kinetic energy?arrow_forward
- 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.)arrow_forwardConsider a frictionless track as shown in Figure P6.62. A block of mass m1 = 5.00 kg is released from . It makes a head-on elastic collision at with a block of mass m2= 10.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision. Figure P6.62arrow_forwardAn bullet of mass m = 8.00 g is fired into a block of mass M = 250 g that is initially at rest at the edge of a table of height h = 1.00 m (Fig. P6.42). The bullet remains in the block, and after the impact the block lands d = 2.00 m from the bottom of the table. Determine the initial speed of the bullet. Figure P6.42arrow_forward
- 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.arrow_forwardConsider a frictionless track as shown in Figure P6.62. A block of mass m1 = 5.00 kg is released from . It makes a head-on elastic collision at with a block of mass m2= 10.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision. Figure P6.62arrow_forwardA car of mass 750 kg traveling at a velocity of 27 m/s in the positive x-direction crashes into the rear of a truck of mass 1 500 kg that is at rest and in neutral at an intersection. If the collision is inelastic and the truck moves forward at 15.0 m/s, what is the velocity of the car after the collision? (See Section 6.3.)arrow_forward
- What is the average momentum of an avalanche that moves a 40-cm-thick layer of snow over an area of 100 m by 500 m over a distance of 1 km down a hill in 5.5 s? Assume a density of 350kg/m3 for the snow.arrow_forwardAn bullet of mass m = 8.00 g is fired into a block of mass M = 250 g that is initially at rest at the edge of a table of height h = 1.00 m (Fig. P6.42). The bullet remains in the block, and after the impact the block lands d = 2.00 m from the bottom of the table. Determine the initial speed of the bullet. Figure P6.42arrow_forwardSuppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is their final velocity?arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningUniversity Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice UniversityPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Glencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-HillCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning