A cart is moving along the $x$-direction with a velocity of $4 \;\mathrm{m/s} $. A person on the cart throws a stone with a velocity of $6 \mathrm{m/s}$ relative to himself. In the frame of reference of the cart the stone is thrown in the $y$-$z$ plane making an angle of $30^\circ$ with the vertical $z$-axis. At the highest point of its trajectory the stone hits an object of equal mass hung vertically from a branch of a tree by means of a string of length $L$. A completely inelastic collision occurs in which the stone gets embedded in the object.

Determine: The speed of the combined mass immediately after the collision with respect to an observer on the ground.

I solved it like this:

The stone is thrown in the $y$-$z$ plane and acceleration along the $z$-axis is $-g$ and that along the $y$-axis is zero.

Momentum is conserved in the $y$-direction and the final velocity immediately after collision is 0 along the $z$-axis and along the $y$-axis the velocity is $V/2$ (from momentum conservation) where $V$ is the velocity along the $y$-axis, which is $6 \cos\left(60\right)=3\mathrm{m/s}$. Hence the final velocity of the combined mass immediately after the collision is $3/2 \mathrm{m/s}= 1.5\mathrm{m/s}$.

Is my solution correct?

Then is there no use for the initial velocity of the cart? Is it independent of the velocity of the combined mass in the ground frame? (The string is assumed to be mass-less.) If my answer is wrong please solve it.


closed as off-topic by user10851, ACuriousMind, Hritik Narayan, user36790, Kyle Kanos Oct 14 '15 at 17:52

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Since the body strikes the other body of same mass with both X and Y components of 4 m/s and 3 m/s respectively, it will collide with an angle . The momentum is conserved in Y direction and X component of velocity remains unaffected. Therefore the velocity of the combined masses will have 4 m/s in X direction and 3/2 m/s in Y direction as you said above,HINT: Take vector sum.


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