Standard +0.3 This is a straightforward energy conservation problem using Hooke's law. Students must equate elastic potential energy at release to kinetic energy plus elastic potential energy at the given position, requiring only standard formula application (EPE = λx²/2l) with no conceptual complications or multi-step reasoning beyond setting up one energy equation.
One end of a light elastic string of natural length \(0.7\) m is attached to a fixed point \(A\) on a smooth horizontal surface. The other end of the string is attached to a particle \(P\) of mass \(0.3\) kg which is held at a point \(B\) on the horizontal surface, where \(AB = 1.2\) m. It is given that \(P\) is released from rest at \(B\) and that when \(AP = 0.9\) m, the particle has speed \(4\) m s\(^{-1}\). Calculate the modulus of elasticity of the string. [3]
Question 1:
1 | λ×0.52 5λ
EE(B) = =
2×0.7 28
λ×0.22 λ
OR EE = =
2×0.7 35
Λ×0.52 λ×0.22 0.3×42
– =
2×0.7 2×0.7 2
λ = 16 N | B1
M1
A1 | [3] | Correct EE when AP = 1.2 m
Correct EE when AP = 0.9 m
Using EE loss = KE gain
One end of a light elastic string of natural length $0.7$ m is attached to a fixed point $A$ on a smooth horizontal surface. The other end of the string is attached to a particle $P$ of mass $0.3$ kg which is held at a point $B$ on the horizontal surface, where $AB = 1.2$ m. It is given that $P$ is released from rest at $B$ and that when $AP = 0.9$ m, the particle has speed $4$ m s$^{-1}$. Calculate the modulus of elasticity of the string. [3]
\hfill \mbox{\textit{CAIE M2 2015 Q1 [3]}}