Video: Elastic Strings and Springs

A ball of mass 1.8 kg is attached to one end of a light elastic string of natural length 2.4 m and modulus of elasticity 17.1 N. The other end of the string is fixed at a point 𝑂. The ball is released from rest at 𝑂. Taking 𝑔 = 9.8 m/sΒ², find how far below 𝑂 the ball reaches before coming instantaneously to rest.

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Video Transcript

A ball of mass 1.8 kilogrammes is attached to one end of a light elastic string of natural length 2.4 metres and modulus of elasticity 17.1 newtons. The other end of the string is fixed at a point 𝑂. The ball is released from rest at 𝑂. Taking 𝑔 equal to 9.8 metres per second squared, find out how far below 𝑂 the ball reaches before coming instantaneously to rest.

We will begin this question by drawing a diagram. The ball is initially released from rest at 𝑂. Until the string reaches its natural length of 2.4 metres, the ball will fall under gravity. We can, therefore, say that the initial velocity 𝑒 is equal to zero and the acceleration is 9.8. We can, therefore, use our equations of motion, or SUVAT equations, to calculate the velocity of the ball when the string reaches its natural length.

𝑠 is equal to 2.4 as this is the distance, or displacement. 𝑒 is equal to zero. π‘Ž is equal to 9.8. And we don’t know the value of 𝑑. The equation that doesn’t contain 𝑑 is 𝑣 squared is equal to 𝑒 squared plus two π‘Žπ‘ . Substituting in our values gives us 𝑣 squared is equal to two multiplied by 9.8 multiplied by 2.4. 𝑣 squared is, therefore, equal to 47.04 metres per second. This means that the velocity 𝑣 is equal to the square root of 47.04.

The ball will then continue to fall until it comes instantaneously to rest. At this point, the velocity will be equal to zero. We will call this distance, which the ball has fallen, π‘₯. This will be the extension in the string. We know that the energies at these two points will be equal. We, therefore, need to consider the kinetic energy, gravitational potential energy, and elastic potential energy at both of these points.

The kinetic energy at any point is equal to a half multiplied by the mass multiplied by the velocity squared. The gravitational potential energy is equal to the mass multiplied by gravity multiplied by the height. Finally, the elastic potential energy is equal to πœ† multiplied by π‘₯ squared divided by two 𝑙. πœ† is the modulus of elasticity, in this case, 17.1 newtons. 𝑙 is the natural length, in this case, 2.4 metres. And π‘₯ is the extension in the string.

We can set the GPE equal to zero at any point. In this case, we’ll choose the point at which the string gets to its natural length. At this point, the elastic potential energy will also be equal to zero as there is no extension in the string. The kinetic energy at this point will be equal to a half multiplied by 1.8, which is the mass, multiplied by 47.04, which is the velocity squared. This is equal to 42.336 joules.

At the point at which the ball comes instantaneously to rest, the kinetic energy will be equal to zero as the velocity is zero. The gravitational potential energy will be equal to 1.8 multiplied by 9.8 multiplied by negative π‘₯. The π‘₯-value is negative as this position is below the previous position. The height is lower. Therefore, our value for π‘₯ is negative. 1.8 multiplied by 9.8 is equal to 17.64. Therefore, our gravitational potential energy is equal to negative 17.64π‘₯.

The elastic potential energy at this point is equal to 17.1 multiplied by π‘₯ squared divided by two multiplied by 2.4. This simplifies to 3.5625π‘₯ squared. As previously mentioned, the energies at both points must be equal. We can, therefore, set up a quadratic equation. 3.5625π‘₯ squared minus 17.64π‘₯ is equal to 42.336. We can set this equal to zero by subtracting 42.336 from both sides. We can then solve the quadratic using the quadratic formula.

Substituting in our values of π‘Ž, 𝑏, and 𝑐 gives us two possible values, π‘₯ is equal to 6.72 or π‘₯ is equal to negative 1.77. As the extension in the string must be positive, we can ignore negative 1.77. The extension in the string when the ball comes instantaneously to rest is 6.72 metres. We might think this is the end of the question. However, we were asked to find out how far below 𝑂 the ball reached. We need to add the natural length 2.4 metres to the extended length of 6.72 metres. We can, therefore, conclude that the ball comes instantaneously to rest 9.12 metres below 𝑂.

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