# Tag Info

1

For inelastic scattering, the initial momentum is $m_b v_{b_i}$. After collision, both $m_b$ and $m_c$ move together, with a velocity $v_{b_f}=v_{c_f}=v_{cm}$. By conservation of momentum $m_b v_{b_i}=m_b v_{b_f}+m_c v_{c_f}=(m_b +m_c)v_{cm}$, whichyield the equation that you are looking for

1

When the man throws the ball, both the ball and the man get equal momentum in the opposite directions. Since the collusion is elastic, i.e: no loss in energy, the ball rebounds with momentum of the same magnitude but in the opposite direction. At this point, both the ball and the man have momentum in the same direction with equal mangitude. When the man ...

1

When a man in frictionless surface throws the ball in forward direction, by conservation of linear momentum he gets pushed back (exactly the case in space where astronaut throws something back to move foreward).Here,when man throws the ball, the momentum of ball and man are exactly equal and their velocities are in opposite direction. But you need to note ...

1

Because your "balls box" runs into problems with propagation delays, etc, I decided to rearrange the problem a little bit - then I think it becomes solvable. In the scenario you want to describe, you have three balls: one with mass $m_0$ moving with velocity $v$ to the right; a second with mass $m_0$ moving with velocity $v$ to the left; and a third with ...

1

When an asteroid collides with the Earth at a speed of, say, 30 km/s, the asteroid will totally disintegrate. The energy involved in the collision between an oxygen atom in the asteroid and an oxygen atom in the Earth is about 75 eV, which is way more than the binding energy. So, the chemical bonds between the atoms in the matter directly involved in the ...

1

Let me switch to vector notation where $v_i$ is the velocity (vector) of ball $i$ prior to collision, and $v^\prime_i$ is the velocity (vector) immediately after collision. When the collision occurs, the two balls impart an equal but opposite impulse on one another of magnitude $J$. The velocities post-collision are then given by \begin{align} v_1^\prime&...

1

Impulse is a vector ($\overrightarrow J= \overrightarrow F \triangle t$) and its absolute value is J. Thus it has two components in this case and the vector can be written $(J_x,J_y)$. where, $$J_x=J\cdot \cos(\theta)=J\cdot \frac{\triangle rx_i + \triangle rx_j}{\sigma_i+sigma_j}$$ And $J_y=J\cdot \sin(\theta)$ can be obtained similarly.

1

You have proved with your analysis that $v_2$ cannot be zero. $m_2$ must be moving with some non-zero velocity, either in the same direction as $m_1$, in which case $v_2$ must be smaller than $v_1$ (or $m_1$ will never catch up to $m_2$) or $m_2$ must be moving in the opposite direction to $m_1$. With the information given, there are 4 unknowns: \$v_1, v_2, ...

Only top voted, non community-wiki answers of a minimum length are eligible