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Chapter 4: Problem 510

A ball is allowed to fall from a height \(20 \mathrm{~m}\). If there is \(30 \%\) loss of energy due to impact, then after one impact ball will go up to (A) \(18 \mathrm{~m}\) (B) \(16 \mathrm{~m}\) (C) \(12 \mathrm{~m}\) (D) \(14 \mathrm{~m}\)

Short Answer

Expert verified
After one impact with a 30% loss of energy, the ball will bounce back up to a height of \(14 \mathrm{~m}\). The answer is (D) \(14 \mathrm{~m}\).

Step by step solution

01

Remember the formula for gravitational potential energy

To solve this problem, we need to recall the formula for gravitational potential energy: \[PE = mgh\] where: - PE is the potential energy, - m is the mass of the ball, - g is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)), - h is the height from the surface.
02

Calculate the initial potential energy of the ball

When the ball is at a height of 20 meters, it has gravitational potential energy, which we can calculate using the formula from step 1. We can ignore the mass as it will cancel out later: \[PE_1 = mgh = mg(20)\]
03

Calculate the potential energy after impact

Since there is a 30% loss of energy during the impact, we need to calculate 70% of the initial potential energy to find the remaining potential energy after impact: \[PE_2 = 0.7 * PE_1 = 0.7 * mg(20)\]
04

Calculate the new height after the impact

Now we need to find the new height the ball will reach after the impact. To do this, we set the new potential energy (PE_2) equal to the gravitational potential energy formula: \[PE_2 = mgh_2\] Since we know the value of PE_2, we can substitute it in and solve for the new height (h_2): \[0.7 * mg(20) = mgh_2\] Divide both sides by mg to get: \[0.7 * 20 = h_2\]
05

Calculate the new height of the ball

Now solve for the new height h_2: \[h_2 = 0.7 * 20 = 14\] So, after one impact with a 30% loss of energy, the ball will bounce back up to a height of 14 meters. The answer is (D) \(14 \mathrm{~m}\).

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Most popular questions from this chapter

A body initially at rest undergoes one dimensional motion with constant acceleration. The power delivered to it at time \(\mathrm{t}\) is proportional to..... (A) \(\mathrm{t}^{1 / 2}\) (B) \(t\) (C) \(\mathrm{t}^{3 / 2}\) (D) \(\mathrm{t}^{2}\)

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