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Sagot :
When the cord carrying the ball is at 30 deg below horizontal, the ball is at
1.5 m - 0.70 sin 30 = 1.15 meters above ground
Energy of the ball at the topmost point of swing :
KE + PE = 1/2 m v² + m g h
= 1/2 0.160 3.26² + 0.160 * 9.81 * (1.50 + 0.70) = 8.767 Joules
Energy of the ball when the cord at 30 deg from horizontal
= 1/2 m V² + m g h =
= 1/2 * 0.160 V² + 0.160 * 9.81 * 1.15 = 0.08V² + 1.805 Joules
Conservation of energy : 0.08 V² + 1.805 = 8.767
V² = 87.025
V = 9.329 m/sec
1.5 m - 0.70 sin 30 = 1.15 meters above ground
Energy of the ball at the topmost point of swing :
KE + PE = 1/2 m v² + m g h
= 1/2 0.160 3.26² + 0.160 * 9.81 * (1.50 + 0.70) = 8.767 Joules
Energy of the ball when the cord at 30 deg from horizontal
= 1/2 m V² + m g h =
= 1/2 * 0.160 V² + 0.160 * 9.81 * 1.15 = 0.08V² + 1.805 Joules
Conservation of energy : 0.08 V² + 1.805 = 8.767
V² = 87.025
V = 9.329 m/sec
Answer:
The speed of the ball is 5.59 m/s.
Explanation:
Given that,
Mass = 0.160 kg
Radius = 70 cm =0.70 m
Distance = 1.50
Speed at top = 3.26 m/s
Angle = 30.0°
We need to calculate the speed of the ball
The total energy at the top
[tex]K.E_{i}+P.E=\dfrac{1}{2}mv_{i}^2+mgh[/tex]
The final kinetic energy of ball at that point when the cord is 30° below the horizontal
[tex]K.E_{f}=\dfrac{1}{2}mv_{f}^{2}[/tex]
Using conservation of energy
[tex]K.E_{f}=K.E_{i}+P.E[/tex]
[tex]\dfrac{1}{2}mv_{f}^2=\dfrac{1}{2}mv_{i}^2+mgh[/tex]
[tex]\dfrac{1}{2}\times0.160\times v_{f}^2=\dfrac{1}{2}\times0.160\times(3.26)^2+0.160\times9.8\times(0.70+0.70\times\sin30^{\circ})[/tex]
[tex]v_{f}=\sqrt{(3.26)^2+2\times9.8\times1.05}[/tex]
[tex]v_{f}=5.59\ m/s[/tex]
Hence, The speed of the ball is 5.59 m/s.
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