Physics, 18.06.2020 20:57 jvsome8

Identical twins, each with mass 61.0 kg, are on ice skates and at rest on a frozen lake, which may be taken as frictionless. Twin A is carrying a backpack of mass 12.0 kg. She throws it horizontally at 3.20 m/s to Twin B. Neglecting any gravity effects, what are the subsequent speeds of Twin A and Twin B

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Immediately after throwing the backpack away, twin A would be moving away from twin B at approximately $0.630\; \rm m \cdot s^{-1}$ .

Initially, twin B would not immediately be moving. However, after the backpack hits her, she would move away from twin A at approximately $0.526\; \rm m \cdot s^{-1}$ if she held onto the backpack.

Explanation:

Consider this scenario in three steps:

Step one: twin A is carrying the backpack.Step two: twin A throws the backpack away; the backpack is en route to twin B;Step three: twin B starts to move after the backpack hits her.

Since all external forces are ignored, momentum should be conserved when changing from step one to step two, and from step two to step three.

From step one to step two

In step one, neither twin A nor the backpack is moving. Their initial momentum would be zero. That is:

$p(\text{twin A, step one}) = 0$ . $p(\text{backpack, step one}) = 0$ .

Therefore:

$p(\text{backpack, step one}) +p(\text{twin A, step one}) = 0$ .

In step two, the backpack is moving towards twin B at $3.20\; \rm m \cdot s^{-1}$ . Since the mass of the backpack is $12.0\; \rm kg$ , its momentum at that point would be:

$\begin{aligned}p(\text{backpack, step two}) &= m \cdot v \\ &= 12.0\;\rm kg \times 3.20\; \rm m \cdot s^{-1} = 38.4\; \rm kg\cdot m \cdot s^{-1} \end{aligned}$ .

Momentum is conserved when twin A throws the backpack away. Hence:

$\begin{aligned}&p(\text{backpack, step two}) +p(\text{twin A, step two}) \\ &= p(\text{backpack, step one}) +p(\text{twin A, step one})\end{aligned}$ .

Therefore:

$p(\text{twin A, step two}) \\ &= p(\text{backpack, step one}) +p(\text{twin A, step one}) - p(\text{backpack, step two}) \\ &= -38.4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}$ .

The mass of twin A (without the backpack) is $61.0\; \rm kg$ . Therefore, her velocity in step two would be:

$\begin{aligned} v(\text{twin A, step two}) &= \frac{p}{m} \\ &= \frac{-38.4\; \rm kg \cdot m \cdot s^{-1}}{61.0\; \rm kg} \approx -0.630\; \rm m \cdot s^{-1}\end{aligned}$ .

Note that while the velocity of the backpack is assumed to be greater than zero, the velocity of twin A here is less than zero. Since the backpack is moving towards twin B, it can be concluded that twin A is moving in the opposite direction away from twin B.

From step two to step three

In step two:

$p(\text{twin B, step two}) = 0$ since twin B is not yet moving. $p(\text{backpack, step two}) = 38.4\; \rm kg \cdot m\cdot s^{-1}$ from previous calculations.

Assume that twin B holds onto the incoming backpack. Thus, the velocity of the backpack and twin B in step three will be the same. Let $v(\text{twin B and backpack, step three})$ denote that velocity.

In step three, the sum of the momentum of twin B and the backpack would thus be:

$\begin{aligned}& m(\text{twin B}) \cdot v(\text{twin B and backpack, step three}) \\ &+ m(\text{backpack}) \cdot v(\text{twin B and backpack, step three})\end{aligned}$ .

Simplify to obtain:

$(m(\text{twin B}) + m(\text{backpack})) \cdot v(\text{twin B and backpack, step three})$ .

Momentum is conserved when twin B receives the backpack. Therefore:

$\begin{aligned}& (m(\text{twin B}) + m(\text{backpack})) \cdot v(\text{twin B and backpack, step three})\\ =&p(\text{twin B, step two}) +p(\text{backpack, step two})\\ =& 38.4\; \rm kg \cdot m\cdot s^{-1} \end{aligned}$ .

Therefore:

$\begin{aligned}& v(\text{twin B and backpack, step three})\\ =&\frac{p(\text{twin B, step two}) +p(\text{backpack, step two})}{m(\text{twin B}) + m(\text{backpack})}\\ =& \frac{38.4\; \rm kg \cdot m\cdot s^{-1}}{61.0\; \rm kg + 12.0\; \rm kg} \approx 0.526\;\rm m\cdot s^{-1} \rm \end{aligned}$ .

In other words, if twin B holds onto the backpack, then (after doing so) she would be moving away from twin A at approximately $0.526\; \rm m \cdot s^{-1}$ .

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