The speed of the thrown ball just before hitting the wall was twice its speed immediately after hitting.
| April 30, 2021 | education
The speed of the thrown ball just before hitting the wall was twice its speed immediately after hitting. Find the kinetic energy of the ball before hitting if an amount of heat equal to 15J was released during the hit.
To find the speed of the ball in question before hitting the wall, we will use the formula: m * V0 ^ 2/2 = Ek0 = Ek1 + Q = m * V1 ^ 2/2 + Q.
Variable values: V0 (speed before hitting the wall) = 2V1 (speed after hitting); Q is the released heat (Q = 15 kJ).
Let’s transform the formula and perform the calculation: m * V0 ^ 2/2 = m * (V0 / 4) ^ 2/2 + 15; m * V0 ^ 2/2 = m * V0 ^ 2 / (2 * 4) + 15; m * V0 ^ 2/2 – m * V0 ^ 2/8 = 3 * m * V0 ^ 2 / (2 * 4) = 15 and Ek0 = m * V0 ^ 2/2 = 15 * 4/3 = 20 J …
Answer: 20 J.
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