Question

Notify Me Bookmark

A 1.5 kg ball is released from the rest from a height 0.409 m. This 1.5 kg ball then swings downward and strikes (fully elastic) a 4.60 kg ball that is at rest. a) Using the principle of conservation of energy, find the speed of the 1.50 kg ball at the height of 0.30 m. b) Find the maximum height of the 4.6 kg ball the collision c) Find the maximum velocity of 4.6 kg ball.

A 1.5 kg ball is released from the rest from a height 0.409 m. This 1.5 kg ball then swings downward and strikes (fully elastic) a 4.60 kg ball that is at rest. a) Using the principle of conservation of energy, find the speed of the 1.50 kg ball at the height of 0.30 m. b) Find the maximum height of the 4.6 kg ball the collision c) Find the maximum velocity of 4.6 kg ball.

Image text
  1. A 1.5 k g ball is released from the rest from a height 0.409 m . This 1.5 k g ball then swings downward and strikes (fully elastic) a 4.60 k g ball that is at rest. a) Using the principle of conservation of energy, find the speed of the 1.50 k g ball at the height of 0.30 m . b) Find the maximum height of the 4.6 k g ball the collision c) Find the maximum velocity of 4.6 k g ball.

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

Starting with an initial speed of 6.00 m/s at a height of 0.321 m, 32.94−kg ball swings downward and strikes a 5.65−kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 2.94−kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.94 kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 5.65−kg ball just after the collision. (d) How high does the 2.94−kg ball swing after the collision, ignoring air resistance? (e) How high does the 5.65−kg ball swing after the collision, ignoring air resistance?
Solution
0 0

Potential energy is given by PE = mgh Kinetic energy is given by KE = mv2 2 Figure 7: Trade in joules between potential and kinetic energy Consider the situation in figure 7, the total mechanical energy is 6 J. There is no loss or gain of mechanical energy, only a transformation from kinetic energy to potential energy (and vice versa). The ball, of mass m = 1 kg is released from rest. Answer the following questions: Question #:1 At point A:What is the potential energy of the ball? What is the height of the ball? What is the velocity of the ball at that point? -What is the kinetic energy? Question #:2 At point B, if the potential energy is equal to the kinetic energy (PE = KE = 3 J)What is the height H of the ball? What is the velocity of the ball at that height? Question #:3 At point C, the potential energy is zero. What is the kinetic energy of the ball? What is the velocity of the ball at that point? How would you describe the velocity at this particular point? Explain Question #:4 At point D, the ball is at the same height as in point A, it reaches a maximum height then stops before reversing direction. What is the potential energy? What is the kinetic energy of the ball? What happened to kinetic energy at that position? Question #:5 While the pendulum is oscillating between positions, use the words increase, decrease, or constant to describe the trend of the different forms of energies. Complete table 2 Table 2: Trend of the different forms of energy of a swinging pendulum Question #:6 Using figure 7, identify the corresponding positions A, B and C in figure 8 Discuss the different energies, and speeds of the glider on those points based on your study of the pendulum.Potential energy is given by PE = mgh Kinetic energy is given by KE = mv2 2 Figure 7: Trade in joules between potential and kinetic energy Consider the situation in figure 7, the total mechanical energy is 6 J. There is no loss or gain of mechanical energy, only a transformation from kinetic energy to potential energy (and vice versa). The ball, of mass m = 1 kg is released from rest. Answer the following questions: Question #:1 At point A:What is the potential energy of the ball? What is the height of the ball? What is the velocity of the ball at that point? -What is the kinetic energy? Question #:2 At point B, if the potential energy is equal to the kinetic energy (PE = KE = 3 J)What is the height H of the ball? What is the velocity of the ball at that height? Question #:3 At point C, the potential energy is zero. What is the kinetic energy of the ball? What is the velocity of the ball at that point? How would you describe the velocity at this particular point? Explain Question #:4 At point D, the ball is at the same height as in point A, it reaches a maximum height then stops before reversing direction. What is the potential energy? What is the kinetic energy of the ball? What happened to kinetic energy at that position? Question #:5 While the pendulum is oscillating between positions, use the words increase, decrease, or constant to describe the trend of the different forms of energies. Complete table 2 Table 2: Trend of the different forms of energy of a swinging pendulum Question #:6 Using figure 7, identify the corresponding positions A, B and C in figure 8 Discuss the different energies, and speeds of the glider on those points based on your study of the pendulum.
Solution
0 0

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.8 kg and 2.4 kg, and the length of the wire is 1.40 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision. (a) v = (b) v =
Solution
0 0