Question 365: Centre of Mass and Rotational Motion

Question

A projectile is launched vertically from the earth with speed $$v_1$$ hits a satellite at the height $$h$$ moving with speed $$v_2$$. If both have the same mass $$m$$, then what is the common velocity if they move together after the collision?

Accepted Answer

Since, momentum remains conserved for a system so,
$$p$$ (before collision) $$=p$$ (after collision)
$$m \mathbf{v}_1+m \mathbf{v}_2=2 m \mathbf{v}_c$$
Where, $$\mathbf{v}_c=$$ common velocity after collision In $$x$$-direction,
$$m v_2=2 m v_{c, x}$$
$$\Rightarrow \quad v_{c, x}=\frac{v_2}{2}$$
Similarly, in $$y$$-direction, $$v_{c, y}=\frac{v_1}{2}$$
$$	herefore$$ Resultant common velocity,
$$\left|\mathbf{v}_cight|=\sqrt{\left(\frac{v_1}{2}ight)^2+\left(\frac{v_2}{2}ight)^2}=\frac{1}{2} \sqrt{v_1^2+v_2^2}$$

Answer

$$v_1+v_2$$

Answer

$$\sqrt{v_1^2+v_2^2}$$

Answer

$$\frac{\sqrt{v_1^2+v_2^2}}{2}$$

Answer

$$2 \sqrt{v_1^2+v_2^2}$$

Centre of Mass and Rotational Motion

A projectile is launched vertically from the earth with speed $v_1 hits a satellite at the height h moving with speed v_2. If both have the same mass m$, then what is the common velocity if they move together after the collision?

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