The depth of the bullet’s penetration into the block.
Answer to Problem 24P
The depth of the bullet’s penetration into the block is
Explanation of Solution
Assume equal firing speeds and equal forces required for the two bullets to push wood fibers apart. This equal forces act backward on two bullets.
Write the expression for law of conservation of energy for first bullet.
Here,
Write the expression for initial kinetic energy of the first bullet.
Here,
Write the expression for mechanical energy of the system.
Here,
Write the expression for law of conservation of momentum for second bullet.
Here,
Rewrite the above expression for mass and velocity.
Here,
Write the expression for law of conservation of energy for second bullet.
Here,
Write the expression for initial kinetic energy of the second bullet.
Here,
Write the expression for mechanical energy of the system.
Here,
Write the expression for final kinetic energy of the second bullet.
Here,
Conclusion:
Substitute the equations (II) and (III) in equation (I).
Substitute
Rewrite the above expression for
Substitute
Substitute the equations (VI), (VII) and (VIII) in equation (V).
Substitute
Substitute equation (XI) in above relation.
Substitute equation (X) in above relation.
Therefore, the depth of the bullet’s penetration into the block is
Want to see more full solutions like this?
Chapter 8 Solutions
Principles of Physics: A Calculus-Based Text
- 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.arrow_forwardGiven the following data for a fire extinguisher-toy wagon rocket experiment, calculate the average exhaust velocity of the gases expelled from the extinguisher. Starting from rest, the final velocity is 10.0 m/s. The total mass is initially 75.0 kg and is 70.0 kg after the extinguisher is fired.arrow_forwardA 100-g firecracker is launched vertically into the air and explodes into two pieces at the peak of its trajectory. If a 72-g piece is projected horizontally to the left at 20 m/s, what is the speed and direction of the other piece?arrow_forward
- An astronaut in her space suit has a total mass of 87.0 kg, including suit and oxygen tank. Her tether line loses its attachment to her spacecraft while shes on a spacewalk. Initially at rest with respect to her spacecraft, she throws her 12.0-kg oxygen tank away from her spacecraft with a speed of 8.00 m/s to propel herself back toward it (Fig. P6.29). (a) Determine the maximum distance she can be from the craft and still return within 2.00 min (the amount of time the air in her helmet remains breathable), (b) Explain in terms of Newtons laws of motion why this strategy works. Figure P6.29arrow_forwardA block with mass m1 = 0.500 kg is released from rest on a frictionless track at a distance h1, = 2.50 m above the top of a table. It then collides elastically with an object having mass m2 = 1.00 kg that is initially at rest on the table, as shown in Figure P6.71. (a) Determine the velocities of the two objects just after the collision. (b) How high up the track does the 0.500-kg object travel back after the collision? (c) How far away from the bottom of the table does the1.00-kg object land, given that the height of the table h2 = 2.00 m? (d) How far away from the bottom of the table does the 0.500-kg object eventually band? Figure P6.71arrow_forwardTwo skateboarders, with masses m1 = 75.0 kg and m2 = 65.0 kg, simultaneously leave the opposite sides of a frictionless half-pipe at height h = 4.00 m as shown in Figure P11.49. Assume the skateboarders undergo a completely elastic head-on collision on the horizontal segment of the half-pipe. Treating the skateboarders as particles and assuming they dont fall off their skateboards, what is the height reached by each skateboarder after the collision? FIGURE P11.49arrow_forward
- A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s an angle of = 60.0 with the surface. It bounces off with the same speed and angle (Fig. P9.67). If the ball is in contact with the wall for 0.200 s, what is the average force exerted by the wall on the ball?arrow_forwardThe mass of the blue puck in Figure P9.44 is 20.0% greater than the mass of the green puck. Before colliding, the pucks approach each other with momenta of equal magnitudes and opposite directions, and the green puck has an initial speed of 10.0 m/s. Find the speeds the pucks have after the collision if half the kinetic energy of the system becomes internal energy during the collision.arrow_forwardA 5.0-g egg falls from a 90-cm-high counter onto the floor and breaks. What impulse is exerted by the floor on the egg?arrow_forward
- A block with mass m1 = 0.500 kg is released from rest on a frictionless track at a distance h1, = 2.50 m above the top of a table. It then collides elastically with an object having mass m2 = 1.00 kg that is initially at rest on the table, as shown in Figure P6.71. (a) Determine the velocities of the two objects just after the collision. (b) How high up the track does the 0.500-kg object travel back after the collision? (c) How far away from the bottom of the table does the1.00-kg object land, given that the height of the table h2 = 2.00 m? (d) How far away from the bottom of the table does the 0.500-kg object eventually band? Figure P6.71arrow_forwardWhat 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?arrow_forwardA 0.500-kg sphere moving with a velocity expressed as (2.00i3.00j+1.00k)m/s strikes a second, lighter sphere of mass 1.50 kg moving with an initial velocity of (1.00i+2.00j3.00k)m/s. (a) The velocity of the 0.500-kg sphere after the collision is (1.00i+3.00j8.00k)m/s. Find the final velocity of the 1.50-kg sphere and identify the kind of collision (elastic, inelastic, or perfectly inelastic). (b) Now assume the velocity of the 0.500-kg sphere after the collision is (0.250i+0.750j2.00k)m/s. Find the final velocity of the 1.50-kg sphere and identify the kind of collision. (c) What If? Take the velocity of the 0.500-kg sphere after the collision as (1.00i+3.00jak)m/s. Find the value of a and the velocity of the 1.50-kg sphere after an elastic collision.arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-HillClassical Dynamics of Particles and SystemsPhysicsISBN:9780534408961Author:Stephen T. Thornton, Jerry B. MarionPublisher:Cengage Learning
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningUniversity Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice UniversityCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College