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30 kg block on spring scale, k = 20 kN/m

Short answer

The block strikes at 6.264 m/s; after plastic impact common speed is 4.698 m/s. Initial spring compression from pan weight is 0.0049 m. Work-energy yields total compression 0.230 m, so pan deflection ≈0.225 m below impact position.

A 30 kg block is dropped from a height of 2 m onto the 10 kg pan of a spring scale. Assuming the impact to be perfectly plastic, determine the maximum deflection of the pan. The constant of the spring is k=20k = 20 kN/m. [Figure: 30 kg block A above 10 kg pan B on a spring, drop height 2 m.]

Data: mA=30m_A=30 kg, mB=10m_B=10 kg, h=2h=2 m, k=20k=20 kN/m, perfectly plastic impact (e=0e=0), g=9.81 m/s2g=9.81\ \text{m/s}^2.

Step 1: block just before impact

vA=2gh=2(9.81)(2)=6.264 m/sv_A=\sqrt{2gh}=\sqrt{2(9.81)(2)}=6.264\ \text{m/s}

Step 2: plastic impact (pan initially at rest)

Momentum is conserved and both move together:

30(6.264)=(30+10)v′ ⇒ v′=4.698 m/s30(6.264)=(30+10)v'\ \Rightarrow\ v'=4.698\ \text{m/s}

Step 3: initial compression of the spring due to the pan

x1=mBgk=10(9.81)20000=0.00490 mx_1=\frac{m_Bg}{k}=\frac{10(9.81)}{20000}=0.00490\ \text{m}

Step 4: work-energy from just after impact to the lowest point

Let x2x_2 be the maximum total compression and M=40M=40 kg. Kinetic energy just after impact: 12Mv′2=441.45\tfrac12Mv'^2=441.45 J.

12Mv′2+12kx12+Mg(x2−x1)=12kx22\tfrac12Mv'^{2}+\tfrac12kx_1^{2}+Mg(x_2-x_1)=\tfrac12kx_2^{2} 10000x22−392.40x2−(441.45+0.2406−1.9247)=010000x_2^{2}-392.40x_2-\left(441.45+0.2406-1.9247\right)=0 x2=0.2302 mx_2=0.2302\ \text{m}

The pan moves down x2−x1=0.2253x_2-x_1=0.2253 m from its position at impact.

Answer: maximum deflection of the pan ≈0.225\approx0.225 m below its position at impact (total spring compression ≈0.230\approx0.230 m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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