Question

Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres eual to 12R. If they attract each other due to gravitational force only , then the distance covered by the smaller body just before collision is
(1)2.5R
(2)4.5R
(3)7.5R
(4)1.5R

Answer 1

Answer on Question#50287 - Physics - Mechanics | Kinematics | Dynamics

(3) 7.5R

Solution

Keep in mind that $x - axis$ actually passes through bodies' centers. Grading in units of $R$ .

$V_{1}$ and $V_{2}$ denote velocities.

Due to Newton's $3^{\mathrm{rd}}$ law:

$$M \frac {d ^ {2} x _ {1}}{d t ^ {2}} = - 5 M \frac {d ^ {2} x _ {2}}{d t ^ {2}}$$

Thus,

$$\frac {d ^ {2} x _ {1}}{d t ^ {2}} = - 5 \frac {d ^ {2} x _ {2}}{d t ^ {2}}$$

, where $x_{1}$ and $x_{2}$ positions of the $1^{\text{st}}$ and the $2^{\text{nd}}$ bodies respectively.

Let us integrate both part with respect to time $t$ .

$$\int \frac {d ^ {2} x _ {1}}{d t ^ {2}} d t = - 5 \int \frac {d ^ {2} x _ {2}}{d t ^ {2}} d t$$ $$\frac {d x _ {1}}{d t} = V _ {1} (t); \frac {d x _ {2}}{d t} = V _ {2} (t);$$ $$\int \frac {d V _ {1}}{d t} d t = - 5 \int \frac {d V _ {2}}{d t} d t$$ $$V _ {1} + C _ {1} = - 5 V _ {2} + C _ {2}$$

, where $C_1$ and $C_2$ - constants of integration.

At initial moment of time $(t = 0)$ both velocities equal to zero, hence $C_1 = C_2 = 0$ .

$$V _ {1} = - 5 V _ {2}$$

Integrate with respect to time once more. Definite integral now. $\tau$ - time at which collision happens.

$$\int_{0}^{\tau} V_{1} dt = -5 \int_{0}^{\tau} V_{2} dt$$ $$\int_{0}^{\tau} \frac{dx_{1}}{dt} dt = -5 \int_{0}^{\tau} \frac{dx_{2}}{dt} dt$$ $$x_{1}(\tau) - x_{1}(0) = -5 \left(x_{2}(\tau) - x_{2}(0)\right)$$

Recall, that due to our choice: $x_{1}(0) = 0$; $x_{2}(0) = 12$;

$$x_{1}(\tau) - 0 = -5 (x_{2}(\tau) - 12)$$ $$x_{1}(\tau) = 60 - 5 x_{2}(\tau)$$

Also, we know that minimal distance between spheres is nothing else, but the sum of its radii.

It means in our case: $x_{2}(\tau) - x_{1}(\tau) = 1 + 2 = 3$.

Thus, we can substitute $x_{2}(\tau) = 3 + x_{1}(\tau)$ into main equation above.

$$x_{1}(\tau) = 60 - 5 \left(3 + x_{1}(\tau)\right)$$ $$x_{1}(\tau) = 60 - 15 - 5 x_{1}(\tau)$$ $$6 x_{1}(\tau) = 45$$ $$x_{1}(\tau) = 7.5$$

Recall that we get result in units of $R$.

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