Question

a mass of 10 g moving with a speed of 100 cm/s strikes a pendulum bob of mass 10 g. the two masses stick together. The maximum height reached by the system now is?

Answer 1

Answer on Question #43697 – Physics - Mechanics | Kinematics | Dynamics

a mass of $10\,\mathrm{g}$ moving with a speed of $100\,\mathrm{cm/s}$ strikes a pendulum bob of mass $10\,\mathrm{g}$. The two masses stick together. The maximum height reached by the system now is?

Solution:

$\mathrm{V}_1 = 1\,\frac{\mathrm{m}}{\mathrm{s}}$ – initial speed of the object;

$\mathrm{m}_1 = 0.01\,\mathrm{kg}$ – mass of the object;

$\mathrm{V}_2 = 0$ – initial speed of pendulum bob;

$\mathrm{m}_2 = 0.01\,\mathrm{kg}$ – mass of the pendulum bob;

$\mathrm{V}_3$ – the speed of the joined masses after the collision;

$\mathrm{H}$ – maximum height reached by the system;

This is an inelastic collision, so momentum is conserved but some energy is lost to heat, etc. in the collision.

Thus:

$$\mathrm{m}_1\,\mathrm{V}_1 + \mathrm{m}_2\,\mathrm{V}_2 = (\mathrm{m}_1 + \mathrm{m}_2)\,\mathrm{V}_3$$ $$\mathrm{V}_3 = \frac{\mathrm{m}_1\,\mathrm{V}_1 + \mathrm{m}_2\,\mathrm{V}_2}{(\mathrm{m}_1 + \mathrm{m}_2)}$$

Now the system (mass + pendulum bob) have a joint KE and they will swing upward until that KE is turned into GPE

$$\mathrm{GPE} = \mathrm{KE}$$ $$(\mathrm{m}_1 + \mathrm{m}_2)\,\mathrm{gH} = \frac{(\mathrm{m}_1 + \mathrm{m}_2)\,\mathrm{V}_3^2}{2}$$

(1) in (2):

$$\mathrm{H} = \frac{\left(\frac{\mathrm{m}_1\,\mathrm{V}_1 + \mathrm{m}_2\,\mathrm{V}_2}{(\mathrm{m}_1 + \mathrm{m}_2)}\right)^2}{2\,\mathrm{g}} = \frac{(\mathrm{m}_1\,\mathrm{V}_1 + \mathrm{m}_2\,\mathrm{V}_2)^2}{2\,\mathrm{g}(\mathrm{m}_1 + \mathrm{m}_2)^2} = \frac{\left(0.1\,\mathrm{kg} \cdot 1\,\frac{\mathrm{m}}{\mathrm{s}} + 0.1\,\mathrm{kg} \cdot 0\right)^2}{2 \cdot 9.8\,\frac{\mathrm{N}}{\mathrm{kg}} \cdot (0.1\,\mathrm{kg} + 0.1\,\mathrm{kg})^2} = 1.28\,\mathrm{cm}$$

Answer: maximum height reached by the system now is $1.28\,\mathrm{cm}$