-1
$\begingroup$

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.

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Community, ACuriousMind, Hritik Narayan, Community, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.