(a) State the triangle law of vector addition. (b) Name the four physical quantities that are associated with the equationq of linear motion. (c) Using the ...
(b) Name the four physical quantities that are associated with the equationq of linear motion.
(c) Using the same set of axes, sketch and label two graphs to illustrate the variation of potential energy and kinetic energy with time for a body in simple harmonic motion.
(d)
A light spiral spring of force constant K lies on a horizontal frictionless surface and has one end fixed to a vertical wall. A block P of mass 2.0 kg placed against the free end of the spring is pushed a distance 5 cm towards the wall with 10J of energy as illustrated in the diagram above. The block is released and after 0.25s, it collides inelastically with a stationary block Q of mass 4.0 kg. Calculate the:
(i) value of k;
(ii) force used to compress the spring;
(iii) acceleration of the block p after release;
(iv) common speed after collision of the blocks.
(a) Triangle law of vector addition
If two vectors are represented, in magnitude and direction, by two sides of a triangle taken in order (head-to-tail), their resultant is represented by the third side of the triangle drawn from the tail of the first vector to the head of the second vector.
This is closely related to the triangle-of-forces statement: three forces in equilibrium can be represented by the three sides of a triangle taken in order.
(b) Physical quantities in the equations of linear motion
The four quantities are:
displacement, \(s\),
initial and final velocity, \(u\) and \(v\),
acceleration, \(a\),
time, \(t\).
For example, \(v=u+at\) relates velocity, acceleration and time, while \(s=ut+\tfrac{1}{2}at^2\) also includes displacement.
(c) Potential energy and kinetic energy in SHM
For an ideal body in simple harmonic motion, total mechanical energy is constant. Kinetic energy is greatest at the equilibrium position, whereas potential energy is greatest at each extreme position. Thus, when one energy is zero, the other is at its maximum value \(E\).
The energy curves have a period of \(T/2\), because energy depends on the square of displacement or velocity. At every instant,
The spring force is greatest at the maximum compression of \(0.050\,\mathrm{m}\):
\[F=kx=(8.0\times10^3)(0.050)=400\,\mathrm{N}.\]
Therefore, the force needed to hold the spring at this compression is \(400\,\mathrm{N}\). The force is not constant while the spring is being compressed; it increases from zero to \(400\,\mathrm{N}\).
(iii) Acceleration of P immediately after release
Immediately after release, the spring force is \(400\,\mathrm{N}\), so Newton’s second law gives
The acceleration is \(200\,\mathrm{m\,s^{-2}}\) immediately after release, directed away from the wall. It then decreases as the spring expands, because \(F=kx\) decreases.
(iv) Common speed after the inelastic collision
The supplied reference solution treats the acceleration as constant for \(0.25\,\mathrm{s}\), producing \(50\,\mathrm{m\,s^{-1}}\). This is not physically consistent: the spring contains only \(10\,\mathrm{J}\), whereas a \(2.0\,\mathrm{kg}\) block moving at \(50\,\mathrm{m\,s^{-1}}\) would have \(2500\,\mathrm{J}\) of kinetic energy.
Since the surface is frictionless, the \(10\,\mathrm{J}\) of elastic potential energy becomes kinetic energy of P when the spring returns to its natural length:
\[\frac{1}{2}m_Pu_P^2=10\]
\[\frac{1}{2}(2.0)u_P^2=10\]
\[u_P=\sqrt{10}=3.16\,\mathrm{m\,s^{-1}}.\]
After leaving the spring, P continues at this constant speed until it reaches Q. Momentum is conserved in the inelastic collision:
\[m_Pu_P+m_Qu_Q=(m_P+m_Q)V.\]
Since Q is stationary, \(u_Q=0\):
\[(2.0)(3.16)+(4.0)(0)=(2.0+4.0)V.\]
\[V=\frac{6.32}{6.0}=1.05\,\mathrm{m\,s^{-1}}.\]
The common speed of the blocks after collision is therefore \(1.05\,\mathrm{m\,s^{-1}}\).
Examination reminder: For a spring, do not use \(v=at\) over the whole motion unless the force, and therefore acceleration, is constant. Here the force falls as the spring expands, so conservation of energy gives the speed correctly.
If two vectors are represented, in magnitude and direction, by two sides of a triangle taken in order (head-to-tail), their resultant is represented by the third side of the triangle drawn from the tail of the first vector to the head of the second vector.
This is closely related to the triangle-of-forces statement: three forces in equilibrium can be represented by the three sides of a triangle taken in order.
(b) Physical quantities in the equations of linear motion
The four quantities are:
displacement, \(s\),
initial and final velocity, \(u\) and \(v\),
acceleration, \(a\),
time, \(t\).
For example, \(v=u+at\) relates velocity, acceleration and time, while \(s=ut+\tfrac{1}{2}at^2\) also includes displacement.
(c) Potential energy and kinetic energy in SHM
For an ideal body in simple harmonic motion, total mechanical energy is constant. Kinetic energy is greatest at the equilibrium position, whereas potential energy is greatest at each extreme position. Thus, when one energy is zero, the other is at its maximum value \(E\).
The energy curves have a period of \(T/2\), because energy depends on the square of displacement or velocity. At every instant,
The spring force is greatest at the maximum compression of \(0.050\,\mathrm{m}\):
\[F=kx=(8.0\times10^3)(0.050)=400\,\mathrm{N}.\]
Therefore, the force needed to hold the spring at this compression is \(400\,\mathrm{N}\). The force is not constant while the spring is being compressed; it increases from zero to \(400\,\mathrm{N}\).
(iii) Acceleration of P immediately after release
Immediately after release, the spring force is \(400\,\mathrm{N}\), so Newton’s second law gives
The acceleration is \(200\,\mathrm{m\,s^{-2}}\) immediately after release, directed away from the wall. It then decreases as the spring expands, because \(F=kx\) decreases.
(iv) Common speed after the inelastic collision
The supplied reference solution treats the acceleration as constant for \(0.25\,\mathrm{s}\), producing \(50\,\mathrm{m\,s^{-1}}\). This is not physically consistent: the spring contains only \(10\,\mathrm{J}\), whereas a \(2.0\,\mathrm{kg}\) block moving at \(50\,\mathrm{m\,s^{-1}}\) would have \(2500\,\mathrm{J}\) of kinetic energy.
Since the surface is frictionless, the \(10\,\mathrm{J}\) of elastic potential energy becomes kinetic energy of P when the spring returns to its natural length:
\[\frac{1}{2}m_Pu_P^2=10\]
\[\frac{1}{2}(2.0)u_P^2=10\]
\[u_P=\sqrt{10}=3.16\,\mathrm{m\,s^{-1}}.\]
After leaving the spring, P continues at this constant speed until it reaches Q. Momentum is conserved in the inelastic collision:
\[m_Pu_P+m_Qu_Q=(m_P+m_Q)V.\]
Since Q is stationary, \(u_Q=0\):
\[(2.0)(3.16)+(4.0)(0)=(2.0+4.0)V.\]
\[V=\frac{6.32}{6.0}=1.05\,\mathrm{m\,s^{-1}}.\]
The common speed of the blocks after collision is therefore \(1.05\,\mathrm{m\,s^{-1}}\).
Examination reminder: For a spring, do not use \(v=at\) over the whole motion unless the force, and therefore acceleration, is constant. Here the force falls as the spring expands, so conservation of energy gives the speed correctly.
