Impulse and perfectly inelastic collision between 2 points

[Thermofox]

Homework Statement

A body, of mass $m_A = 2kg$, is initially at rest on the ground. $A$ is then subjected to a vertical instantaneous impulse of magnitude $J = 10 Ns$. When $A$ has a velocity of $v_A = 5m/s$, the body is struck in a perfectly inelastic collision by the body B. Knowing that $m_B=0.2Kg$ and that $B$ has a constant horizontal velocity, $v_B= 5m/s$, determine:
1) $h$
2) The energy lost in the collision
3) $\theta$
4) The maximum height of $A+B$ after the collision

Relevant Equations

$\Delta P = I$

The main thing about this problem is to find the components of the velocity, $v_{A+B}$. To do that you have to use the conservation of linear momentum of the collision. In this case, since there is an impulse, I should have $\Delta P = J$. But the impulse is given prior to the collision. Does this mean that $\Delta P = 0$ in the collision? That's the thing I don't understand.

After I figure that out, I know how to finish the problem:
1)I Already can determine $h$ by writing an energy balance from when $A$ is on the ground to the moment right before the collision with $B$;
2) Energy lost = $|\Delta E_{kinetic}|$ ;
3) Once I have the components of $v_{A+B}$, I can determine $\theta$ with trigonometry;
4) Max height, $h_{max}= \frac {v_{A+B}^2 (\sin\theta)^2} {2g}$.

 

[erobz]

Homework Statement: A body, of mass $m_A = 2kg$, is initially at rest on the ground. $A$ is then subjected to a vertical instantaneous impulse of magnitude $J = 10 Ns$. When $A$ has a velocity of $v_A = 5m/s$, the body is struck in a perfectly inelastic collision by the body B. Knowing that $m_B=0.2Kg$ and that $B$ has a constant horizontal velocity, $v_B= 5m/s$, determine:
1) $h$
2) The energy lost in the collision
3) $\theta$
4) The maximum height of $A+B$ after the collision
Relevant Equations: $\Delta P = I$

View attachment 347751The main thing about this problem is to find the components of the velocity, $v_{A+B}$. To do that you have to use the conservation of linear momentum of the collision. In this case, since there is an impulse, I should have $\Delta P = J$. But the impulse is given prior to the collision. Does this mean that $\Delta P = 0$ in the collision? That's the thing I don't understand.

After I figure that out, I know how to finish the problem:
1)I Already can determine $h$ by writing an energy balance from when $A$ is on the ground to the moment right before the collision with $B$;
2) Energy lost = $|\Delta E_{kinetic}|$ ;
3) Once I have the components of $v_{A+B}$, I can determine $\theta$ with trigonometry;
4) Max height, $h_{max}= \frac {v_{A+B}^2 (\sin\theta)^2} {2g}$.

The impulse $J$ they are talking about is applied only to mass $m_A$ (before the collision).

So yeah, if you are saying gravity (impulse) is negligibly small over the very short collision duration ($A$ impacting $B$) then for the collision $\Delta P = 0$.

 

[Thermofox]

The impulse $J$ they are talking about is applied only to mass $m_A$ (before the collision).

So yeah, if you are saying gravity is negligibly small over the very short collision duration ($A$ impacting $B$) then for the collision $\Delta P = 0$.

Ok thanks for the clarification.

 

[erobz]

Ok thanks for the clarification.

Also, don't forget that you are starting at $h$, when finding $h_{max}$.

 

[Thermofox]

Also, don't forget that you are starting at $h$, when finding $h_{max}$.

Right, $h_{max}= \frac {v_{A+B}^2 (\sin\theta)^2} {2g}$ is valid only when $h_{\text{initial}}=0$.
$\Rightarrow h_{max}= h_{\text{initial}} + \frac {v_{A+B}^2 (\sin\theta)^2} {2g}$