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Fynjy0 [20]
4 years ago
8

A 1400-kg car traveling east at 25m/s collides with a 1800-kg car traveling at a speed of 20m/s in a direction that makes angle

of 40degree south of west. the cars stick together after the collision. what is the magnitude and direction of the velocity of the cars after the collision?
Physics
1 answer:
julia-pushkina [17]4 years ago
7 0
M1*V1 + M2*V2 = M1*V + M2*V.
1400*25 + 1800*20[180+40]=1400*V+1800*V.
Divide both sides by 100:
14*25 + 18*20[220o] = 14V + 18V.
350 + 360[220o] = 32V.
350 - 276-231i = 32V.
74 - 231i = 32V.
242.6[-72.2o] = 32V.
V = 7.6m/s[-72.2o]=7.6m/s[72o] S. of E.
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Answer:

\boxed {\boxed {\sf 2.1 \ kilograms}}

Explanation:

Kinetic energy is the energy an object possesses due to motion. The formula 1/2 the product of mass and the squared velocity.

E_k=\frac{1}{2} mv^2

We know the baseball's kinetic energy is 105 Joules. It is also traveling at a velocity of 10 meters per second. `

First, convert the units of Joules to make unit cancellation easier later in the problem. 1 Joule (J) is equal to 1 kilogram square meter per square second (kg*m²/s²). The baseball's kinetic energy of 105 J is equal to 105 kg*m²/s².

Now we know 2 values:

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Substitute these values into the formula.

105 \ kg*m^2/s^2= \frac{1}{2} m (10 \ m/s)^2

Now we need to solve for m, the mass. Solve the exponent.

  • (10 m/s)²= 10 m/s * 10 m/s = 100 m²/s²

105 \ kg *m^2/s^2 = \frac{1}{2} m (100 \ m^2/s^2)

Multiply on the right side.

105 \ kg *m^2/s^2 =  m (\frac{1}{2} * 100 \ m^2/s^2)

105 \ kg *m^2/s^2 =  m (50 \ m^2/s^2)

The variable, m, is being multiplied by 50 square meters per square second. The opposite of multiplication is division, so we divide both sides by that value.

\frac {105 \ kg *m^2/s^2 }{50 \ m^2/s^2}=  \frac{ m (50 \ m^2/s^2)}{50 \ m^2/s^2}

\frac {105 \ kg *m^2/s^2 }{50 \ m^2/s^2}= m

The units of square meter per square second will cancel out.

\frac {105 }{50} \ kg= m

2.1 \ kg=m

The mass of the baseball is <u>2.1 kilograms. </u>

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A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds
aleksklad [387]

<em>There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem</em>

Answer:

(a) \Delta P=1.67 \ kg.m/s

(b) \Delta P=0.86\ m/s

Explanation:

<u>Change of Momentum</u>

The momentum of a given particle of mass m traveling at a speed v is given by

P=m.v

When this particle changes its speed to a value v', the new momentum is

P'=m.v'

The change of momentum is

\Delta P=m.v'-m.v

\Delta P=m.(v'-v)

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J

This potential energy is transformed into kinetic energy just before the collision occurs, thus

\displaystyle \frac{1}{2}mv^2=6.0368

Solving for v

\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}

v=-14\ m/s

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

h'=88.5\% (10)=8.85\ m

The potential energy reached by the ball at its rebound is

U'=m.g.h'=0.0616\cdot 9.8\cdot 8.85 =5.342568\ J

Thus the speed after the collision is

\displaystyle v'=\sqrt{\frac{5.342568\cdot 2}{0.0616}}

v'=13.17\ m/s

The change of momentum is

\Delta P=0.0616\cdot (13.17+14)

\Delta P=1.67 \ kg.m/s

(b) If the putty sticks to the floor, then v'=0

\Delta P=0.0616\cdot (0+14)

\Delta P=0.86\ m/s

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4 years ago
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