Question

1. (a) a ball has a mass of 0.25 kg. a student holds the ball 1.75 m above the ground. (i) state the equation linking gravitational potential energy (gpe), mass, g and height. (ii) calculate the gravitational potential energy of the ball. gpe = ………… j (3 marks) (b) the student lets the ball fall. state the value of the kinetic energy (ke) of the ball just before it hits the ground. assume that there is no air resistance. ke = ………… j (1 mark) (c) another ball with the same mass has a kinetic energy of 3.1 j. (i) state the equation linking kinetic energy, mass and speed. (ii) calculate the speed of the ball. speed = ………… m/s

Explanation:

Step1: State gravitational - potential - energy formula

The formula for gravitational potential energy is $GPE = mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity ($g = 9.8\ m/s^{2}$), and $h$ is the height.

Step2: Calculate gravitational potential energy

Given $m = 0.25\ kg$, $g=9.8\ m/s^{2}$, and $h = 1.75\ m$. Substitute these values into the formula:
$GPE=mgh=0.25\times9.8\times1.75 = 4.2875\ J$

Step3: Use energy - conservation for falling ball

When the ball falls and there is no air resistance, the gravitational potential energy at the start is converted entirely into kinetic energy just before it hits the ground. So, $KE = GPE=4.2875\ J$

Step4: State kinetic - energy formula

The formula for kinetic energy is $KE=\frac{1}{2}mv^{2}$, where $m$ is the mass and $v$ is the speed.

Step5: Calculate speed of the ball

Given $KE = 3.1\ J$ and $m = 0.25\ kg$. Rearrange the kinetic - energy formula $v=\sqrt{\frac{2KE}{m}}$.
Substitute the values: $v=\sqrt{\frac{2\times3.1}{0.25}}=\sqrt{24.8}\approx4.98\ m/s$

Answer:

(a)(i) $GPE = mgh$
(a)(ii) $4.29\ J$ (rounded to 2 decimal places)
(b) $4.29\ J$ (rounded to 2 decimal places)
(c)(i) $KE=\frac{1}{2}mv^{2}$
(c)(ii) $4.98\ m/s$ (rounded to 2 decimal places)