(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 ...

(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,

\[KE+PE=E.\]

(d) Spring, block and collision

Given:

• mass of block P: \(m_P=2.0\,\mathrm{kg}\),
• mass of block Q: \(m_Q=4.0\,\mathrm{kg}\),
• compression: \(x=5.0\,\mathrm{cm}=0.050\,\mathrm{m}\),
• elastic potential energy stored: \(E=10\,\mathrm{J}\).

(i) Force constant of the spring

The elastic potential energy in a compressed spring is

\[E=\frac{1}{2}kx^2.\]

Therefore,

\[10=\frac{1}{2}k(0.050)^2\]

\[k=\frac{2(10)}{(0.050)^2}=8.0\times10^3\,\mathrm{N\,m^{-1}}.\]

(ii) Force used to compress the spring

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

\[a=\frac{F}{m_P}=\frac{400}{2.0}=200\,\mathrm{m\,s^{-2}}.\]

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.