Question:
3. In this question, two spheres are supported by strings. When the two spheres are pulled back and released, they collide and rebound at a position as shown below.
The table below provides data for the two spheres X and Y during this collision. (Convention: the positive direction is horizontal and it is to the right.)
(a) By using the law of conservation of linear momentum to determine the mass of sphere Y. [3]
(b) State and explain whether the collision of the two spheres is elastic. [1]
(c) Use Newton’s second law of motion and Newton’s third law of motion to explain why the magnitudes of the change in momentum of both spheres have the same value. [3]
4. A spring is attached to a movable slider at A and a fixed support at B as shown below. A cart (m = 1.7 kg) moves with a velocity v towards the slider. The cart collides with the slider and the spring is compressed such that they are momentarily at rest.
A graph of compression x of the spring is varied with force F exerted on the spring is shown below.
(a) The cart and the spring are momentarily at rest when the restoring force (F) of the spring is 4.5 N.
(i) Show that the restoring force (compression) of the spring follows Hooke’s law. [2]
(ii) Calculate the elastic constant (or spring constant, k) of the spring. [2]
(iii) Determine the elastic potential energy (EP) stored in the spring due to the cart when it is momentarily at rest. [3]
(b) Calculate the speed of the cart immediately after it collides with the slider. You may assume that the kinetic energy of the cart (after the collision) is completely converted to the elastic potential energy of the spring. [2]
Mark Scheme:
3(a) Left hand side: 50 × (+4.5) + M × (– 2.8) [1].
Right hand side: 50 × (–1.8 ) + M × (+1.4) [1].
M = 75 g [1].
(b) The relative speed of approach does not equal to the relative speed of separation. Inelastic collision [1].
(c) Newton’s third law of motion: the magnitude of force on X is equal to the magnitude of force on Y [1].
Newton’s second law of motion: force is equal to the rate of change of momentum [1].
The time of collision is the same for both balls [1].
4 (a) (i) The spring constant is determined by two sets of coordinates [1]
F / x is shown to be constant [1]
Alternative: the gradient is calculated by using at least one point on the graph [1]
Show that the y-intercept is zero [1].
(ii) The gradient is calculated, e.g., from 4.5 / 1.8 × 10–2 [1].
The spring constant is 250 N m–1 [1].
(iii) work done = ½Fx or ½kx2 [1]
Ep = ½Fx = 0.5 × 4.5 × 1.8 × 10–2 or Ep = ½kx2 = 0.5 × 250 × (1.8 × 10–2)2 [1]
Energy stored = 0.0405 J [1].
Possible answers:
3(a) Using principle of conservation of linear momentum, muX + MuY = mvX + MvY.
We have 50 × (+4.5) + M × (– 2.8) = 50 × (–1.8 ) + M × (+1.4)
Thus, M = 75 g = 0. 075 kg.
(b) For elastic collision, the relative speed of approach is equal to the relative speed of separation.
However, (uY – uX) = (–2.8 – [+4.5]) = –7.3 and (vY – vX) = (+1.4 –[–1.8]) = +3.2.
That is, the magnitudes of (uY – uX) and (vY – vX) are not equal.
Thus, the collision is inelastic.
(c) By using Newton’s third law of motion, the magnitude of force on X is equal to the magnitude of force on Y, but they are in opposite direction (FX= –FY).
By using Newton’s second law of motion, force is equal to the rate of change of momentum. Thus, FX = ΔpX/ΔtX = –FY = –ΔpY/ΔtY).
Since the time of collision is the same for both balls, hence the change in momentum is the same. (i.e. ΔtX = ΔtY implies ΔpX = –ΔpY or |ΔpX| = |ΔpY|)
4 (a) (i) The straight line graph can be represented by y = mx + c.
(b) KE = ½mv2 = 0.0405
Thus, v = (2× KE/m)1/2 = [2 × 0.0405 / 1.7]1/2 =0.218 m s–1
Feynman’s insights and goofs:
In his lectures, Feynman uses terms such as “elastic,” “effectively perfectly elastic,” “very nearly elastic,” and “various degrees of elasticity.” For example, he states that “[i]n these circumstances the velocities of rebound are practically equal to the initial velocities; such a collision is called elastic (Feynman et al., 1963, section 10–4 Momentum and energy).” In addition, Feynman mentions that “when equilibrium has set in, the net result is that the collisions are effectively perfectly elastic (Feynman et al., 1963, section 39–2 The pressure of a gas).” However, inelastic collision is sometimes defined as a collision in which the coefficient of restitution is between 0 and 1. (A definition of coefficient of restitution, e, is the ratio of the relative velocity after a collision to relative velocity before the collision.) Thus, it is potentially confusing that a collision having a coefficient of restitution of 0.9 may be described as inelastic, nearly elastic, or practically elastic.
On the other hand, Feynman explains that “[f]or a large number of materials, experiments show that for sufficiently small extensions the force is proportional to the extension F ∝ Δl. This relation is known as Hooke’s law (Feynman et al., 1964, section 38–1 Hooke’s law).” In general, Hooke’s law is also known as an empirical law because it is formulated by experiments. An important experimental condition for Hooke’s law is that it is applicable to an elastic spring or string when it is not stretched or compressed beyond the elastic limit. In other words, this empirical law is a good approximation of the elastic behavior of a substance when the deformations are relatively small. Mathematically, Hooke’s law is more precisely expressed as F = -kx because the direction of the restoring force exerted by the spring on an object is opposite to the displacement of the object. Strictly speaking, a real spring does not obey this law beyond a limited (linear) range.
References:
1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
2. Feynman, R. P., Leighton, R. B., & Sands, M. (1964). The Feynman Lectures on Physics, Vol II: Mainly electromagnetism and matter. Reading, MA: Addison-Wesley.
3. In this question, two spheres are supported by strings. When the two spheres are pulled back and released, they collide and rebound at a position as shown below.
The table below provides data for the two spheres X and Y during this collision. (Convention: the positive direction is horizontal and it is to the right.)
(a) By using the law of conservation of linear momentum to determine the mass of sphere Y. [3]
(b) State and explain whether the collision of the two spheres is elastic. [1]
(c) Use Newton’s second law of motion and Newton’s third law of motion to explain why the magnitudes of the change in momentum of both spheres have the same value. [3]
4. A spring is attached to a movable slider at A and a fixed support at B as shown below. A cart (m = 1.7 kg) moves with a velocity v towards the slider. The cart collides with the slider and the spring is compressed such that they are momentarily at rest.
A graph of compression x of the spring is varied with force F exerted on the spring is shown below.
(i) Show that the restoring force (compression) of the spring follows Hooke’s law. [2]
(ii) Calculate the elastic constant (or spring constant, k) of the spring. [2]
(iii) Determine the elastic potential energy (EP) stored in the spring due to the cart when it is momentarily at rest. [3]
(b) Calculate the speed of the cart immediately after it collides with the slider. You may assume that the kinetic energy of the cart (after the collision) is completely converted to the elastic potential energy of the spring. [2]
Mark Scheme:
3(a) Left hand side: 50 × (+4.5) + M × (– 2.8) [1].
Right hand side: 50 × (–1.8 ) + M × (+1.4) [1].
M = 75 g [1].
(b) The relative speed of approach does not equal to the relative speed of separation. Inelastic collision [1].
(c) Newton’s third law of motion: the magnitude of force on X is equal to the magnitude of force on Y [1].
Newton’s second law of motion: force is equal to the rate of change of momentum [1].
The time of collision is the same for both balls [1].
4 (a) (i) The spring constant is determined by two sets of coordinates [1]
F / x is shown to be constant [1]
Alternative: the gradient is calculated by using at least one point on the graph [1]
Show that the y-intercept is zero [1].
(ii) The gradient is calculated, e.g., from 4.5 / 1.8 × 10–2 [1].
The spring constant is 250 N m–1 [1].
(iii) work done = ½Fx or ½kx2 [1]
Ep = ½Fx = 0.5 × 4.5 × 1.8 × 10–2 or Ep = ½kx2 = 0.5 × 250 × (1.8 × 10–2)2 [1]
Energy stored = 0.0405 J [1].
(b) ½mv2 = 0.0405 [1] v = 0.218 m s–1 [1]
Possible answers:
3(a) Using principle of conservation of linear momentum, muX + MuY = mvX + MvY.
We have 50 × (+4.5) + M × (– 2.8) = 50 × (–1.8 ) + M × (+1.4)
Thus, M = 75 g = 0. 075 kg.
(b) For elastic collision, the relative speed of approach is equal to the relative speed of separation.
However, (uY – uX) = (–2.8 – [+4.5]) = –7.3 and (vY – vX) = (+1.4 –[–1.8]) = +3.2.
That is, the magnitudes of (uY – uX) and (vY – vX) are not equal.
Thus, the collision is inelastic.
(c) By using Newton’s third law of motion, the magnitude of force on X is equal to the magnitude of force on Y, but they are in opposite direction (FX= –FY).
By using Newton’s second law of motion, force is equal to the rate of change of momentum. Thus, FX = ΔpX/ΔtX = –FY = –ΔpY/ΔtY).
Since the time of collision is the same for both balls, hence the change in momentum is the same. (i.e. ΔtX = ΔtY implies ΔpX = –ΔpY or |ΔpX| = |ΔpY|)
4 (a) (i) The straight line graph can be represented by y = mx + c.
The gradient can be calculated by using two coordinates:
m = (y2 – y1)/(x2 – x1) = (4.5 – 1.5)/(0.018 – 0.006) = 250
The y-intercept can be calculated as follows:
c = y – mx = 1.5 – (250)(0.006) = 0
The graph (F = kx) is a straight line having a constant gradient and it passes through the origin. Thus, it obeys Hooke’s law.
(ii) From the graph, when x1 = 1.8 cm and F1 = 4.5 N, k = 4.5/0.018 = 250 N/m.
Alternatively, we may use x2 = 0.6 cm and F2 = 1.5 N, k = 1.5/0.006 = 250 N/m.
(iii) The elastic potential energy EP stored in the spring = ½kx2 = 0.5 × 250 × (0.018)2 = 0.0405 J
Thus, v = (2× KE/m)1/2 = [2 × 0.0405 / 1.7]1/2 =0.218 m s–1
Feynman’s insights and goofs:
In his lectures, Feynman uses terms such as “elastic,” “effectively perfectly elastic,” “very nearly elastic,” and “various degrees of elasticity.” For example, he states that “[i]n these circumstances the velocities of rebound are practically equal to the initial velocities; such a collision is called elastic (Feynman et al., 1963, section 10–4 Momentum and energy).” In addition, Feynman mentions that “when equilibrium has set in, the net result is that the collisions are effectively perfectly elastic (Feynman et al., 1963, section 39–2 The pressure of a gas).” However, inelastic collision is sometimes defined as a collision in which the coefficient of restitution is between 0 and 1. (A definition of coefficient of restitution, e, is the ratio of the relative velocity after a collision to relative velocity before the collision.) Thus, it is potentially confusing that a collision having a coefficient of restitution of 0.9 may be described as inelastic, nearly elastic, or practically elastic.
On the other hand, Feynman explains that “[f]or a large number of materials, experiments show that for sufficiently small extensions the force is proportional to the extension F ∝ Δl. This relation is known as Hooke’s law (Feynman et al., 1964, section 38–1 Hooke’s law).” In general, Hooke’s law is also known as an empirical law because it is formulated by experiments. An important experimental condition for Hooke’s law is that it is applicable to an elastic spring or string when it is not stretched or compressed beyond the elastic limit. In other words, this empirical law is a good approximation of the elastic behavior of a substance when the deformations are relatively small. Mathematically, Hooke’s law is more precisely expressed as F = -kx because the direction of the restoring force exerted by the spring on an object is opposite to the displacement of the object. Strictly speaking, a real spring does not obey this law beyond a limited (linear) range.
References:
1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
2. Feynman, R. P., Leighton, R. B., & Sands, M. (1964). The Feynman Lectures on Physics, Vol II: Mainly electromagnetism and matter. Reading, MA: Addison-Wesley.
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