Determine the time taken for the blocks to come into contact with each other.

As a result of the collision, the blocks reverse their direction of motion and travel at the same speed as each other. During the collision, 20% of the kinetic energy of the blocks is given off as thermal energy to the surroundings.

(i) State and explain whether the collision is elastic or inelastic.

(ii) Show that the final speed of the blocks relative to the surface is 0.16 m s^–1.

(i) State Newton’s third law of motion.

(ii) During the collision of the blocks, the magnitude of the force that block A exerts on block B is F_AB and the magnitude of the force that block B exerts on block A is F_BA. On the diagram below, draw labelled arrows to represent the magnitude and direction of the forces F_AB and F_BA.

(iii) The duration of the collision between the blocks is 0.070 s. Determine the average force one block exerted on the other.

distance between surfaces of blocks=0.900-0.050=0.850m;
relative speed between blocks = 0.36ms^–1;
\({\rm{time}} = \frac{{{\rm{distance}}}}{{{\rm{speed}}}} = \frac{{0.850}}{{0.36}} = 2.4{\rm{s}}\);

or

blocks moving at same speed so meet at mid-point;
distance travelled by block=0.450-0.025=0.425m;
\({\rm{time}} = \frac{{{\rm{distance}}}}{{{\rm{speed}}}} = \frac{{0.425}}{{0.18}} = 2.4{\rm{s}}\);

Award [3] for bald correct answer.
Award [2 max] if distance of 0.90 m or 0.45 m used to get 2.5 s.

(i) the collision is inelastic;
because kinetic energy is not conserved (although momentum is);

(ii) initial \({E_K} = \frac{1}{2} \times 0.17 \times {0.18^2} = 0.002754{\rm{J}}\);
final \({E_K} = 0.80 \times 0.002754 = 0.0022032{\rm{J}}\);
final speed \( = \sqrt {\frac{{2 \times 0.0022032}}{{0.17}}} \);
=0.16ms^-1

or

0.8×initial E_K=final E_K;
\(0.8 \times \frac{1}{2} \times 0.17 \times {0.18^2} = \frac{1}{2} \times 0.17 \times {v^2}\);
\(v = \sqrt {0.8 \times {{0.18}^2}} \);
=0.16ms^-1

(i) if object A exerts a force on object B, then object B (simultaneously) exerts an equal and opposite force on object A / every action has an equal and opposite reaction / OWTTE;

(ii) arrows of equal length; (judge by eye)
acting through centre of blocks;
correct labelling consistent with correct direction;

(iii) \(\Delta v = 0.16 - \left( { - 0.18} \right) = 0.34{\rm{m}}{{\rm{s}}^{ - 1}}\);
\(a = \frac{{\Delta v}}{{\Delta t}} = \frac{{0.34}}{{0.070}} = 4.857{\rm{m}}{{\rm{s}}^{ - 2}}\);
\(F = ma = 0.17 \times 4.857 = 0.83{\rm{N}}\);

or

\(\Delta v = 0.16 - \left( { - 0.18} \right) = 0.34{\rm{m}}{{\rm{s}}^{ - 1}}\);
\({\rm{impulse}} = F\Delta t = m\Delta v \Rightarrow F = \frac{{0.17 \times 0.34}}{{0.07}}\);
F=0.83N;