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3 years ago

15

In a very busy off-campus eatery one chef sends a 235-g broccoli-tomato-pickle-onion-mushroom pizza sliding down the counter fro

m left to right at 1.65 m/s. Almost simultaneously, the other chef launches a 351-g veggieburger-on-a-bun (with everything, but hold the horseradish) along the same counter from right to left at 2.47 m/s. The two delicacies collide head on at the given speeds–the counter is practically friction-free due to accumulated grease–and merge into a single, unbelievably savory serving. In what direction does the delectible dish move, if it moves at all?

1 answer:

STALIN [3.7K]3 years ago

3 0

Answer:

0.47922 kgm/s and will move right to left

Explanation:

$m_1$ = Mass of broccoli-tomato-pickle-onion-mushroom pizza = 235 g

$m_2$ = Mass of veggieburger-on-a-bun = 351 g

$v_1$ = Velocity of broccoli-tomato-pickle-onion-mushroom pizza = 1.65 m/s

$v_2$ = Velocty of veggieburger-on-a-bun = -2.47 m/s

Left to right is considered positive direction and the opposite is negative direction

Net momentum is

$p=m_1v_1+m_2v_2\\\Rightarrow p=0.235\times 1.65+0.351\times -2.47\\\Rightarrow p=-0.47922\ kgm/s$

The combined dish has a momentum of 0.47922 kgm/s and will move right to left

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Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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Answer:

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