Question
A bullet of mass M1 is fired towards a block
of mass m2 initially at rest at the edge of a
frictionless table of height h as in the figure.
The initial speed of the bullet is vi. Consider
two cases, a completely inelastic one and an
elastic one,where the bullet bounces off the
block.
What is the ratio of the flight time; i.e.,
tAB/tA′B′ ?
of mass m2 initially at rest at the edge of a
frictionless table of height h as in the figure.
The initial speed of the bullet is vi. Consider
two cases, a completely inelastic one and an
elastic one,where the bullet bounces off the
block.
What is the ratio of the flight time; i.e.,
tAB/tA′B′ ?
Answers
GPT-4o mini
To determine the ratio of the flight times \( t_{AB} / t_{A'B'} \) for the two scenarios (completely inelastic collision where the bullet embeds in the block, and elastic collision where the bullet bounces off), we need to analyze the motions in the vertical direction, where both cases involve the same fall height \( h \).
### 1. Case 1: Completely Inelastic Collision
In a completely inelastic collision, the bullet of mass \( M_1 \) collides with the block of mass \( M_2 \) and they move together after the collision.
#### Flight Time Calculation (t_AB):
1. **Before the collision:**
The bullet is fired horizontally toward the edge of the table with an initial speed \( v_i \) and falls a vertical distance \( h \) when it reaches the edge.
2. **After the collision:**
The bullet and block move together with a new velocity \( V_f \) found using conservation of momentum:
\[
(M_1 + M_2) V_f = M_1 v_i \implies V_f = \frac{M_1 v_i}{M_1 + M_2}
\]
3. **Flight time during free fall:**
The time \( t_{AB} \) it takes for the bullet-block system to hit the ground depends only on the vertical motion:
\[
h = \frac{1}{2} g t_{AB}^2
\]
Solving for \( t_{AB} \):
\[
t_{AB} = \sqrt{\frac{2h}{g}}
\]
### 2. Case 2: Elastic Collision
In the elastic collision, the bullet bounces off the block. Therefore, we first determine the speed of the bullet and the block after the collision.
#### Flight Time Calculation (t_A′B′):
1. **Before the collision:**
The bullet is again fired horizontally with an initial speed \( v_i \).
2. **After the collision:**
For an elastic collision, we can use the following equations for one-dimensional elastic collisions:
The velocities after the collision are given by:
- Bullet \( v_{1f} = \frac{M_1 - M_2}{M_1 + M_2} v_i \)
- Block \( v_{2f} = \frac{2M_1}{M_1 + M_2} v_i \)
However, for the flight time calculation, we focus on the vertical motion.
3. **Flight time during free fall:**
Since the vertical motion is independent of the horizontal motion, the free fall distance \( h \) is the same as in the inelastic case. Thus:
\[
h = \frac{1}{2} g t_{A'B'}^2
\]
Solving for \( t_{A'B'} \):
\[
t_{A'B'} = \sqrt{\frac{2h}{g}}
\]
### Conclusion: Ratio of Flight Times
Since both \( t_{AB} \) and \( t_{A'B'} \) depend only on the height \( h \) and \( g \), we see that:
\[
t_{AB} = t_{A'B'} = \sqrt{\frac{2h}{g}}
\]
Thus, the ratio of the flight times is:
\[
\frac{t_{AB}}{t_{A'B'}} = \frac{\sqrt{\frac{2h}{g}}}{\sqrt{\frac{2h}{g}}} = 1
\]
Therefore, the ratio of the flight times for the two cases is:
\[
\boxed{1}
\]
### 1. Case 1: Completely Inelastic Collision
In a completely inelastic collision, the bullet of mass \( M_1 \) collides with the block of mass \( M_2 \) and they move together after the collision.
#### Flight Time Calculation (t_AB):
1. **Before the collision:**
The bullet is fired horizontally toward the edge of the table with an initial speed \( v_i \) and falls a vertical distance \( h \) when it reaches the edge.
2. **After the collision:**
The bullet and block move together with a new velocity \( V_f \) found using conservation of momentum:
\[
(M_1 + M_2) V_f = M_1 v_i \implies V_f = \frac{M_1 v_i}{M_1 + M_2}
\]
3. **Flight time during free fall:**
The time \( t_{AB} \) it takes for the bullet-block system to hit the ground depends only on the vertical motion:
\[
h = \frac{1}{2} g t_{AB}^2
\]
Solving for \( t_{AB} \):
\[
t_{AB} = \sqrt{\frac{2h}{g}}
\]
### 2. Case 2: Elastic Collision
In the elastic collision, the bullet bounces off the block. Therefore, we first determine the speed of the bullet and the block after the collision.
#### Flight Time Calculation (t_A′B′):
1. **Before the collision:**
The bullet is again fired horizontally with an initial speed \( v_i \).
2. **After the collision:**
For an elastic collision, we can use the following equations for one-dimensional elastic collisions:
The velocities after the collision are given by:
- Bullet \( v_{1f} = \frac{M_1 - M_2}{M_1 + M_2} v_i \)
- Block \( v_{2f} = \frac{2M_1}{M_1 + M_2} v_i \)
However, for the flight time calculation, we focus on the vertical motion.
3. **Flight time during free fall:**
Since the vertical motion is independent of the horizontal motion, the free fall distance \( h \) is the same as in the inelastic case. Thus:
\[
h = \frac{1}{2} g t_{A'B'}^2
\]
Solving for \( t_{A'B'} \):
\[
t_{A'B'} = \sqrt{\frac{2h}{g}}
\]
### Conclusion: Ratio of Flight Times
Since both \( t_{AB} \) and \( t_{A'B'} \) depend only on the height \( h \) and \( g \), we see that:
\[
t_{AB} = t_{A'B'} = \sqrt{\frac{2h}{g}}
\]
Thus, the ratio of the flight times is:
\[
\frac{t_{AB}}{t_{A'B'}} = \frac{\sqrt{\frac{2h}{g}}}{\sqrt{\frac{2h}{g}}} = 1
\]
Therefore, the ratio of the flight times for the two cases is:
\[
\boxed{1}
\]