Types of impact; collar on curved rod

Impact and its types

Impact is the collision of two bodies during a very short time interval, with large forces acting. The common normal at the contact point is the line of impact.

• Central impact: centres of mass lie on the line of impact. It is direct (velocities along the line) or oblique (velocities inclined to it).
• Eccentric impact: the line of impact misses the centre of mass of a body.
• By $e$ (coefficient of restitution): perfectly elastic $e=1$, perfectly plastic $e=0$, partly elastic $0<e<1$.

Collar problem

Motion is in a horizontal plane, so weight does no work. The normal force from the smooth rod does no work. Only the spring does work, so

$$\tfrac12mv_1^{2}+\tfrac12k\delta_1^{2}=\tfrac12mv_2^{2}+\tfrac12k\delta_2^{2}$$

Assumptions (figure not legible):

• $k=400$ N/m (the value $400$ kN/m would make the collar speed unrealistically large).
• The spring's fixed end D is such that the spring lengths are $L_A=150$ mm, $L_B=\sqrt{150^{2}+200^{2}}=250$ mm and $L_C=100$ mm.
• Undeformed length $L_0=80$ mm, so $\delta_A=0.070$ m, $\delta_B=0.170$ m, $\delta_C=0.020$ m. $m=0.5$ kg.

(i) Speed at A so that the collar just reaches B

$$\tfrac12(0.5)v_A^{2}+\tfrac12(400)(0.070)^{2}=0+\tfrac12(400)(0.170)^{2}$$ $$0.25v_A^{2}=5.78-0.98=4.80\ \Rightarrow\ v_A=4.38\ \text{m/s}$$

(ii) Speed at C

From B (at rest) to C:

$$0+\tfrac12(400)(0.170)^{2}=\tfrac12(0.5)v_C^{2}+\tfrac12(400)(0.020)^{2}$$ $$5.78=0.25v_C^{2}+0.08\ \Rightarrow\ v_C=4.77\ \text{m/s}$$

Answer: (i) $v_A\approx4.38$ m/s; (ii) $v_C\approx4.77$ m/s (for the assumed geometry; the method is the energy balance above with the lengths read from the figure).