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The double ball drop problem is as follows:

A ball of mass $m$ is placed on top of a ball of mass $M$ (where $m < M$), and the balls are dropped simultaneously from some height $h$. When the balls hit the floor, the ball on top will shoot upwards. What is the velocity of this ball at the moment it shoots upwards?

And so I ask; is there any easy derivation of this problem, without the use of calculus?

I have attempted to solve it used the conservation of momentum, kinetic energy, and mechanical energy, but I seem unable to solve it. My attempt is as follows:

First, we can solve for the velocity of the larger ball using the conservation of momentum:

$$p_\text{before} = p_\text{after}$$

$$(m + M)\sqrt{2gh} = mv_m + Mv_M$$

$$v_M = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

Then, using the conservation of kinetic energy:

$$\frac{1}{2}(m + M)\sqrt{2gh}^2 = \frac{1}{2}mv_m^2 + \frac{1}{2}Mv_M^2$$

$$v_M = \sqrt{\frac{2gh(m + M) - mv_m^2}{M}}$$

And then you can equate the two to get:

$$\sqrt{\frac{2gh(m + M) - mv_m^2}{M}} = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

However, this, when solved, gives $v_m = \sqrt{2gh}$, which does not align with what I have been taught (saying that the velocity should be about three times this).

What am I doing wrong in this derivation? The math is correct, I'm sure. It's somewhere in the physics itself, a concept that I'm missing.

Any help is appreciated.

The double ball drop problem is as follows:

A ball of mass $m$ is placed on top of a ball of mass $M$ (where $m < M$), and the balls are dropped simultaneously from some height $h$. When the balls hit the floor, the ball on top will shoot upwards. What is the velocity of this ball at the moment it shoots upwards?

And so I ask; is there any easy derivation of this problem, without the use of calculus?

I have attempted to solve it used the conservation of momentum, kinetic energy, and mechanical energy, but I seem unable to solve it. My attempt is as follows:

First, we can solve for the velocity of the larger ball using the conservation of momentum:

$$p_\text{before} = p_\text{after}$$

$$(m + M)\sqrt{2gh} = mv_m + Mv_M$$

$$v_M = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

Then, using the conservation of kinetic energy:

$$\frac{1}{2}(m + M)\sqrt{2gh}^2 = \frac{1}{2}mv_m^2 + \frac{1}{2}Mv_M^2$$

$$v_M = \sqrt{\frac{2gh(m + M) - mv_m^2}{M}}$$

And then you can equate the two to get:

$$\sqrt{\frac{2gh(m + M) - mv_m^2}{M}} = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

However, this, when solved, gives $v_m = \sqrt{2gh}$, which does not align with what I have been taught (saying that the velocity should be about three times this).

What am I doing wrong in this derivation? The math is correct, I'm sure. It's somewhere in the physics itself, a concept that I'm missing.

Any help is appreciated.

The double ball drop problem is as follows:

A ball of mass $m$ is placed on top of a ball of mass $M$ (where $m < M$), and the balls are dropped simultaneously from some height $h$. When the balls hit the floor, the ball on top will shoot upwards. What is the velocity of this ball at the moment it shoots upwards?

And so I ask; is there any easy derivation of this problem, without the use of calculus?

I have attempted to solve it used the conservation of momentum, kinetic energy, and mechanical energy, but I seem unable to solve it. My attempt is as follows:

First, we can solve for the velocity of the larger ball using the conservation of momentum:

$$p_\text{before} = p_\text{after}$$

$$(m + M)\sqrt{2gh} = mv_m + Mv_M$$

$$v_M = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

Then, using the conservation of kinetic energy:

$$\frac{1}{2}(m + M)\sqrt{2gh}^2 = \frac{1}{2}mv_m^2 + \frac{1}{2}Mv_M^2$$

$$v_M = \sqrt{\frac{2gh(m + M) - mv_m^2}{M}}$$

And then you can equate the two to get:

$$\sqrt{\frac{2gh(m + M) - mv_m^2}{M}} = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

However, this, when solved, gives $v_m = \sqrt{2gh}$, which does not align with what I have been taught (saying that the velocity should be about three times this).

What am I doing wrong in this derivation? The math is correct, I'm sure. It's somewhere in the physics itself, a concept that I'm missing.

Any help is appreciated.

The double ball drop problem is as follows:

A ball of mass $m$ is placed on top of a ball of mass $M$ (where $m < M$), and the balls are dropped simultaneously from some height $h$. When the balls hit the floor, the ball on top will shoot upwards. What is the velocity of this ball at the moment it shoots upwards?

And so I ask; is there any easy derivation of this problem, without the use of calculus?

I have attempted to solve it used the conservation of momentum, kinetic energy, and mechanical energy, but I seem unable to solve it. My attempt is as follows:

First, we can solve for the velocity of the larger ball using the conservation of momentum:

$$p_\text{before} = p_\text{after}$$

$$(m + M)\sqrt{2gh} = mv_m + Mv_M$$

$$v_M = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

Then, using the conservation of kinetic energy:

$$\frac{1}{2}(m + M)\sqrt{2gh}^2 = \frac{1}{2}mv_m^2 + \frac{1}{2}Mv_M^2$$

$$v_M = \sqrt{\frac{2gh(m + M) - mv_m^2}{M}}$$

And then you can equate the two to get:

$$\sqrt{\frac{2gh(m + M) - mv_m^2}{M}} = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

However, this, when solved, gives $v_m = \sqrt{2gh}$, which does not align with what I have been taught (saying that the velocity should be about three times this).

What am I doing wrong in this derivation? The math is correct, I'm sure. It's somewhere in the physics itself, a concept that I'm missing.

Any help is appreciated.

The double ball drop problem is as follows:

A ball of mass $m$ is placed on top of a ball of mass $M$ (where $m < M$), and the balls are dropped simultaneously from some height $h$. When the balls hit the floor, the ball on top will shoot upwards. What is the velocity of this ball at the moment it shoots upwards?

Here is a picture of accompany this description:

And so I ask; is there any easy derivation of this problem, without the use of calculus?

I have attempted to solve it used the conservation of momentum, kinetic energy, and mechanical energy, but I seem unable to solve it. My attempt is as follows:

First, we can solve for the velocity of the larger ball using the conservation of momentum:

$p_{before} = p_{after}$

$(m + M)\sqrt{2gh} = mv_m + Mv_M$

$v_M = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$

Then, using the conservation of kinetic energy:

$\frac{1}{2}(m + M)\sqrt{2gh}^2 = \frac{1}{2}mv_m^2 + \frac{1}{2}Mv_M^2$

$v_M = \sqrt{\frac{2gh(m + M) - mv_m^2}{M}}$

And then you can equate the two to get:

$\sqrt{\frac{2gh(m + M) - mv_m^2}{M}} = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$

However, this, when solved, gives $v_m = \sqrt{2gh}$, which does not align with what I have been taught (saying that the velocity should be about three times this).

What am I doing wrong in this derivation? The math is correct, I'm sure. It's somewhere in the physics itself, a concept that I'm missing.

Any help is appreciated.

The double ball drop problem is as follows:

A ball of mass $m$ is placed on top of a ball of mass $M$ (where $m < M$), and the balls are dropped simultaneously from some height $h$. When the balls hit the floor, the ball on top will shoot upwards. What is the velocity of this ball at the moment it shoots upwards?

And so I ask; is there any easy derivation of this problem, without the use of calculus?

I have attempted to solve it used the conservation of momentum, kinetic energy, and mechanical energy, but I seem unable to solve it. My attempt is as follows:

First, we can solve for the velocity of the larger ball using the conservation of momentum:

$$p_\text{before} = p_\text{after}$$

$$(m + M)\sqrt{2gh} = mv_m + Mv_M$$

$$v_M = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

Then, using the conservation of kinetic energy:

$$\frac{1}{2}(m + M)\sqrt{2gh}^2 = \frac{1}{2}mv_m^2 + \frac{1}{2}Mv_M^2$$

$$v_M = \sqrt{\frac{2gh(m + M) - mv_m^2}{M}}$$

And then you can equate the two to get:

$$\sqrt{\frac{2gh(m + M) - mv_m^2}{M}} = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

However, this, when solved, gives $v_m = \sqrt{2gh}$, which does not align with what I have been taught (saying that the velocity should be about three times this).

What am I doing wrong in this derivation? The math is correct, I'm sure. It's somewhere in the physics itself, a concept that I'm missing.

Any help is appreciated.