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geniusboy [140]
3 years ago
14

Anthony and Sissy are participating in the "Roll-a-Rama" rollerskating dance championship. While 75.6 kg Anthony rollerskates ba

ckwards at 3.7 m/s, 59 kg Sissy jumps into his arms with a velocity of 5.4 m/s in the same direction. How fast does the pair roll backwards together? Answer to the nearest tenth of a m/s.
Physics
1 answer:
hammer [34]3 years ago
8 0

Answer:

4.4 m/s

Explanation:

momentum is always conserved so we can use conversation of momentum to solve the question, also momentum is a vector quantity ( it has magnitude and direction) which is the product of the bodies mass and velocity.

conservation law of momentum relates by the formula below:

momentum before collision = momentum after collision

M1U1 + M2U2 = M1V1 + M2V2

in the case of this two, the formula becomes

M1U1 + M2U2 = V (M1 + M2) since she jumped into his arm

there masses are M1 = 75.6 kg M2 = 59 kg and their velocities are  U1 = 3.7 m/s and U2 = 5.4 m/s, their common velocity after collision = V since their motion is backward the formula becomes

-M1U1 - M2U2 = V(M1 + M2)

substitute the values into the equations

(-75.6 × 3.7 ) + (- 59 × 5.4) = V ( 75.6 + 59)

- 598.32 = 134.6 V

divide both side by 134.6

V = - 598.32 / 134.6 = -4.445 m/s  = -4.4 m/s to nearest tenth the negative means in the same backward direction

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A lemming running 2.22m/s runs off a horizontal cliff. It lands in the water 7.45m from the base of the cliff. How much time was
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6 0
3 years ago
An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1
motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment M_{l}=2.90\times10^{-28}\ kg

Mass of the heavier fragment M_{h}=1.62\times10^{-27}\ Kg

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}

\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}

\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}

\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}

\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c

\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2

\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}

\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}

\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}

\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})

\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}

v_{2}^2=\dfrac{c^2}{8.93}

v_{2}=0.335c

Hence, The speed of the heavier fragment is 0.335c.

7 0
3 years ago
If a roller coaster car had 40,000 J of gravitational potential energy when at rest on the top of a hill how much kinetic energy
Inessa [10]

Answer:

K.E = 30,000 J

Explanation:

Given,

The potential energy of the roller coaster car, P.E = 40000 J

The kinetic energy at height h/4, K.E = ?

According to the law of conservation of energy, the total energy of the system is conserved.

At height 'h', the total energy is,

                                    P.E = mgh

                                     K.E = 0

At height 'h/4', the total energy is

                                     P.E + K.E = mgh

                                     P.E = mgh/4

                                     K.E = 1/2 mv²

Therefore,

                                   mgh/4 + 1/2 mv² = mgh

                                    gh/4 + v²/2 = gh

Hence,

                                      v² = 3gh/2

Substituting in the K.E equation

                               K.E = 1/2 mv²

                                      = 1/2 m (3gh/2)

                                       = 3/4 mgh

                                        = 3/4 x 40000

                                         = 30000 J

Hence, the K.E of the roller coaster car is, K.E = 30000 J

6 0
3 years ago
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