Answer to Question #143811 in Classical Mechanics for isa villa

Question

Upon reaching its maximum height, a 546-g projectile launched from the ground with an initial velocity of 15.0 m/s, 30.0° above the horizontal collides with a vertical free falling object of mass 245 g and the two stick together after the collision. Suppose that at the time of collision, the second object is momentarily at rest, what is the resulting speed of the two objects upon reaching the ground?

Answer 1

As per the given question,

Mass of the object $(M)=546g =0.546kg$

Initial velocity of the projectile v= 15.0 m/sec

Angle of projection $(\theta)=30^\circ$

Mass of the free falling object $(m) =245g$

As the free falling object is momentarily at rest, it means velocity of the object at that instance =0

Now, taking vertical and horizontal component of the velocity, for the projectile

$v_x=v\cos\theta = 15\cos(30^\circ) =12.99 m/s$

$v_y=v\sin\theta =15\sin(30^\circ) =7.5 m/s$

So, here momentum along the horizontal direction, will be conserve

As here, after the collision second object gets stick into the first object and let's final velocity of the object becomes $V_1$ .

Hence $0.456\times 12.99 = (0.245+0.456)\times V_1$

$V_1=\frac{0.456\times 12.99}{0.701} m/sec$

$\Rightarrow V_1=8.45 m/s$

As here, same force is working on the object in the vertical direction, so there will be no change in the speed of the object in the vertical direction, so the speed of the object at the time of the collision to the ground will remains unchanged.

Hence final velocity of the object $V=\sqrt{V_1^2+v_y^2} =\sqrt{8.45^2+7.5^2} m/s$

Hence, the resulting speed of the two objects upon reaching the ground is $V=11.3 m/s$