Knowee
Questions
Features
Study Tools

You are watching a YouTube video to prepare for a physics exam. The 'teacher' in the video states that a ball is elastic because it deforms when it collides with a wall or floor. Are they correct? Carefully explain your answer, and include definitions of deformation and elasticity. (b) [7 marks]A 25.0 cm tall spring is compressed by 12 mm when a 1.00 kg mass is placed on it.(i) (1 mark)Determine the spring constant of the spring.(ii) (1 mark)Calculate the elastic potential energy stored in the spring.(iii) (1 mark)Where did the energy stored in the spring 'come from'?(iv) (2 marks)The spring is now placed horizontally on a frictionless surface with one end attached to a wall and a 740 g mass attached to the other end. The mass is given a push to set it oscillating. What is the frequency of oscillation?(v) (2 marks)A pendulum is placed above the mass-spring system and the bob oscillates back and forth exactly in time with the mass. How long is the pendulum?

Question

You are watching a YouTube video to prepare for a physics exam. The 'teacher' in the video states that a ball is elastic because it deforms when it collides with a wall or floor. Are they correct? Carefully explain your answer, and include definitions of deformation and elasticity. (b) [7 marks]A 25.0 cm tall spring is compressed by 12 mm when a 1.00 kg mass is placed on it.(i) (1 mark)Determine the spring constant of the spring.(ii) (1 mark)Calculate the elastic potential energy stored in the spring.(iii) (1 mark)Where did the energy stored in the spring 'come from'?(iv) (2 marks)The spring is now placed horizontally on a frictionless surface with one end attached to a wall and a 740 g mass attached to the other end. The mass is given a push to set it oscillating. What is the frequency of oscillation?(v) (2 marks)A pendulum is placed above the mass-spring system and the bob oscillates back and forth exactly in time with the mass. How long is the pendulum?

...expand
🧐 Not the exact question you are looking for?Go ask a question

Solution

The 'teacher' in the video is partially correct. When a ball collides with a wall or floor, it does deform, meaning it changes shape. However, this alone does not make the ball elastic. Elasticity refers to the ability of an object to return to its original shape after being deformed. So, if the ball returns to its original shape after the collision, then it can be considered elastic.

(i) The spring constant (k) can be determined using Hooke's Law, which states that the force (F) exerted by a spring is equal to the displacement (x) times the spring constant (k). In this case, the force is the weight of the 1.00 kg mass, which is 9.8 N (1 kg * 9.8 m/s^2), and the displacement is 12 mm or 0.012 m. So, k = F/x = 9.8 N / 0.012 m = 816.67 N/m.

(ii) The elastic potential energy (U) stored in the spring can be calculated using the formula U = 1/2 * k * x^2. Substituting the values, we get U = 1/2 * 816.67 N/m * (0.012 m)^2 = 0.059 Nm or 0.059 J.

(iii) The energy stored in the spring came from the work done to compress the spring, which is equal to the force applied times the displacement.

(iv) The frequency of oscillation (f) can be calculated using the formula f = 1/(2π) * sqrt(k/m), where m is the mass of the object. Substituting the values, we get f = 1/(2π) * sqrt(816.67 N/m / 0.74 kg) = 1.36 Hz.

(v) The period of oscillation (T) of the pendulum is equal to the period of the mass-spring system, which is the reciprocal of the frequency. So, T = 1/f = 1/1.36 s = 0.735 s. The length (L) of the pendulum can be calculated using the formula T = 2π * sqrt(L/g), where g is the acceleration due to gravity. Solving for L, we get L = (T^2 * g) / (4π^2) = (0.735 s)^2 * 9.8 m/s^2 / (4π^2) = 0.44 m.

This problem has been solved

Similar Questions

A 25.0 cm tall spring is compressed by 12 mm when a 1.00 kg mass is placed on it.(i) (1 mark)Determine the spring constant of the spring.(ii) (1 mark)Calculate the elastic potential energy stored in the spring.(iii) (1 mark)Where did the energy stored in the spring 'come from'?(iv) (2 marks)The spring is now placed horizontally on a frictionless surface with one end attached to a wall and a 740 g mass attached to the other end. The mass is given a push to set it oscillating. What is the frequency of oscillation?(v) (2 marks)A pendulum is placed above the mass-spring system and the bob oscillates back and forth exactly in time with the mass. How long is the pendulum?

Calculate the spring constant of the ball when it has an elastic potential store of 0.54 J, after being compressed by 0.23m._____ N/mGive your answer to 2 significant figures.

1. Prince stretched a circular spring with a spring constant of 3 N/m until it was extended to 50cm. What is the elastic potential energy of the spring stretched by Prince/ stored by the spring?2. How much elastic potential energy does a spring store when it is compressed by 0.2 m if ithas a spring constant of 5 N/m?3. What is the elastic potential energy stored in a spring whose spring constant is 160 N/m when itis 8.0 cm compressed?

A 20kg mass is placed on a large spring. A Force of 2000N is applied downwards to compress the spring by 35cm. a. Determine the spring’s force constant. b. Find the elastic potential energy stored in the spring. c. If the spring is released, calculate the kinetic energy of the stone when it returns to its original position. d. Assuming no loss of energy, determine the maximum height that the mass will reach

Which of the following statements is incorrect?All objects will undergo inelastic deformation if the forces acting on them are large enoughWhen an object is deformed elastically, it will return to its original shape if the forces acting on it are removedEnergy is not required to deform an object elasticallyWhen an object is deformed inelastically, it will not return to its original shape if the forces acting on it are removed2What is meant by the limit of proportionality of a spring?A spring may not be stretched beyond its limit of proportionalityWhen a spring is stretched beyond its limit of proportionality, it will not return to its original shape if the force being applied to it is removedIt is the point beyond which the extension of the spring is no longer proportional to the force applied to itOn an extension-force graph for the spring, it is the point after which the line will become straight3Calculate the force which must be applied to extend a spring of spring constant 10 N/m by 5 cm.  You may assume that this force does not cause the spring to exceed its limit of proportionality.0.5 N20 N0.2 N5 N4A spring is initially 30 cm long.  When a force of 12 N is applied to it, its length increases to 50 cm.  Calculate its spring constant.  You may assume that this force does not cause the spring to exceed its limit of proportionality.40 N/m24 N/m60 N/m2.4 N/m5A student is investigating how the force applied to two different springs affects their extension.  She plots an extension-force graph for spring A (with extension on the y-axis and force on the x-axis) and obtains a straight line.  On the same axes, she plots a second straight line using the data which she obtained for spring B.  The gradient (slope) of the line for spring B is lower than that for spring A.  What does this tell us about the spring constants of these springs?The spring constant of spring A is equal to that of spring BNothingThe spring constant of spring A is lower than that of spring BThe spring constant of spring A is greater than that of spring B6A spring (of spring constant 20 N/m) is stretched by 5 cm.  Calculate the amount of work which must be done in stretching the spring by this amount.0.025 J1 J100 J0.5 J

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.