40 kg block on spring scale, central impact

Central impact and its types

Central impact occurs when the centres of mass of the two colliding bodies lie on the line of impact (the common normal at the contact point).

• Direct central impact: both velocities are along the line of impact.
• Oblique central impact: one or both velocities are inclined to the line of impact; the velocity components along the line of impact follow the impact laws, and the tangential components stay unchanged (smooth bodies).

By the coefficient of restitution $e$:

Maximum deflection of the pan

Data: $m_A=40$ kg, $m_B=10$ kg, $h=1.5$ m, $k=25$ kN/m, $e=0$ (perfectly plastic), $g=9.81$ m/s$^2$.

Step 1: block just before impact

$$v_A=\sqrt{2gh}=\sqrt{2(9.81)(1.5)}=5.425\ \text{m/s}$$

Step 2: plastic impact (pan initially at rest, momentum conserved):

$$40(5.425)=(40+10)v'\ \Rightarrow\ v'=4.340\ \text{m/s}$$

Step 3: initial spring compression (pan alone in equilibrium):

$$x_1=\frac{m_Bg}{k}=\frac{10(9.81)}{25000}=0.003924\ \text{m}$$

Step 4: work-energy from just after impact to the lowest point (total compression $x_2$, mass $M=50$ kg):

$$\tfrac12 M v'^{2}+\tfrac12kx_1^{2}+Mg(x_2-x_1)=\tfrac12kx_2^{2}$$

With $\tfrac12Mv'^2=470.88$ J, this gives

$$12500x_2^{2}-490.5x_2-(470.88+0.1925-1.9247)=0$$ $$12500x_2^{2}-490.5x_2-469.15=0\ \Rightarrow\ x_2=0.2143\ \text{m}$$

The pan therefore moves down $x_2-x_1=0.2143-0.0039=0.2104$ m from its position at impact.

Answer: maximum deflection of the pan from its position at impact $\approx0.210$ m (total spring compression $\approx0.214$ m).