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krek1111 [17]
2 years ago
11

Your friend (68 kg) is wearing frictionless roller skates and standing still. You throw at her a 3.6 kg pumpkin with a velocity

of 9.5 m/s and she catches it. What will be her resultant velocity ?
Physics
1 answer:
Fudgin [204]2 years ago
6 0

Answer:

Approximately 0.48\; \rm m\cdot s^{-1}.

Explanation:

Momentum would be conserved since there's no friction on this friend, and all other forces on her are balanced. Therefore:

\begin{aligned}& \text{Resultant momentum of (friend and pumpkin)} \\ =\; & \text{Initial momentum of (friend)} \\ & + \text{Initial momentum of (pumpkin)}\end{aligned}.

Momentum p the product of mass m and velocity v. That is: p = m \, v.

The initial momentum of this friend is 0\; \rm kg \cdot m \cdot s^{-1} since she was initially not moving (an initial velocity of 0\; \rm m\cdot s^{-1}.)

The initial momentum of the pumpkin would be:

\begin{aligned}p &= m \, v \\ &= 3.6 \; \rm kg \times 9.5\; \rm m\cdot s^{-1} \\ &= 34.2\; \rm kg \cdot m \cdot s^{-1}\end{aligned}.

Therefore:

\begin{aligned}& \text{Resultant momentum of (friend and pumpkin)} \\ =\; & \text{Initial momentum of (friend)} \\ & + \text{Initial momentum of (pumpkin)} \\ =\; &0\; {\rm kg \cdot m \cdot s^{-1}} + 34.2\; {\rm kg \cdot m \cdot s^{-1}} \\ =\; & 34.2\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}.

Rearrange the equation p = m \, v to find an expression for velocity v given momentum and mass:

\displaystyle v = \frac{p}{m}.

Note that the "final momentum of friend and pumpkin" in the previous equation refers to the resultant velocity of the friend with the pumpkin in her hand. Thus, it would necessary to use the combined mass of the friend and the pumpkin (68\; {\rm kg} + 3.6 \; {\rm kg}) when calculating the resultant velocity:

\begin{aligned}& \text{Resultant velocity of (friend and pumpkin)} \\ =\; & \frac{\text{Resultant momentum of (friend and pumpkin)}}{\text{Mass of (friend and pumpkin)}} \\ =\; & \frac{34.2\; {\rm kg \cdot m \cdot s^{-1}}}{68\; {\rm kg} + 3.6\; {\rm kg}} \\ \approx \; & 0.48\; \rm m \cdot s^{-1}\end{aligned}.

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Explanation:

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Explanation:

First, denote our known values;

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Mass is impulse divided by change in velocity:

m=\frac{J}{v-u}\\=\frac{1.3}{5.4--3.8}\\\\=\frac{1.3}{9.2}\\=0.1413Kgs

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If a force of 32000N exerted pressure of 160N/m² , find the area on which the force acts.​
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A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the
yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

       x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

       x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

        x = ½ a₁ t²

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Let's solve

       a₁  t² = a₂ (t-4)²

      a₁/a₂ t² = t² -2 4 t + 16

      t² (1- 2.0 / 4.0) - 8 t +16

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Let's solve the second degree equation

     t = [16 ±√( 16² - 4 32)] / 2  

     t = ½ (16 ± 11,3)

Solutions

     t1 = 13.66 s

     t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

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