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3 years ago

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2. Two toy cars are involved in a race. Car A has mass m while car B has mass 2m. a. The two cars have the same force applied to

them over a distance of 1 meter. Which car has a larger kinetic energy after traveling 1 meter? Which car has a larger momentum after traveling 1 meter? Explain your answers. b. The two cars have the same force applied to them over a time period of 10 seconds. Which car has a larger kinetic energy after 10 seconds? Which car has a larger momentum after 10 seconds? Explain your answers.

1 answer:

ArbitrLikvidat [17]3 years ago

7 0

Answer:

a) The kinetic energy of the two cars is the same

the moment of car 2 is greater than the moment of car 1

b) the kinetic energy of car 1 is greater than that of car 2

the moment of the two cars is the same

Explanation:

a) to know the kinetic energy of each car, we must find the speed, use Newton's second law to find the acceleration

Car 1

F = m a

a = F / m

Let's use kinematics to find the velocity after x = 1 m

v² = v₀² + 2 a x

The initial speed is zero

v = √ (2 F/m x)

For the distance of x = 1 m

v₁ = √ (2 F / m)

Car 2

F = 2m a

a = F / 2m

v² = 2 a x

v = √ (F/m x)

For x = 1 m

v₂ = √(F / m)

Let's calculate the kinetic energy of each car

Car 1

K₁ = ½ m v₁²

K₁ = ½ m 2F / m

K₁ = F

Car 2

K₂ = ½ 2m v₂²

K₂ = ½ 2m F / m

K₂ = F

The kinetic energy of the two cars is the same

Let's calculate the moment

Car 1

P₁ = m v₁

P₁ = m √ (2F / m)

Car 2

P₂ = 2m v²

P₂ = 2m √(F / m)

We see that the moment of car 2 is greater than the moment of car 1

b) in this part the force is applied by t = 10 s

Acceleration is the same, let's find the speed

Car1

v = v₀ + a t

v = F / m t

v₁ = F / m 10

Car 2

v₂ = F / 2m 10

v₂ = F / m 5

Let's calculate the kinetic energy of each car

Car 1

K₁ = ½ m v₁²

K₁ = ½ m (F / m 10)²

K₁ = 50 F² / m

Car2

K₂ = ½ 2m v₂²

K₂ = m (F / m 5)²

K₂ = 25 F² / m

In this case we see that the kinetic energy of car 1 is greater than that of car 2

Let's calculate the moment

Car 1

P₁ = m v₁

P₁ = m F / m 10

P₁ = 10 F

Car 2

P₂ = 2m v₂

P₂ = 2m F / m 5

P₂ = 10 F

In this case the moment of the two cars is the same

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There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = $\frac{1}{2}$ m. $v^{2}$

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

$\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f} )$

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

$v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i} + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}$

$v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}$

For all the collisions, object 2 is static, i.e. $v_{2i}$ = 0

<u>Part A</u>: Both objects have the same mass (m), $v_{1i}$ = v and collision is elastic:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = 0

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.m}{m+m}.v$

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

<u>Part B</u>: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

$m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}$

$v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m.v}{m+m}$

v_1 = v_2 = $\frac{v}{2}$

<u>Part C:</u> Object 1 is 2m, object 2 is m and elastic collision:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = $\frac{2m - m}{2m + m } . v$

v_1 = $\frac{v}{3}$

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.2m}{2m+m}.v$

v_2 = $\frac{4v}{3}$

<u>Part D</u>: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = $v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m}{m+3m}.v$

v_1 = v_2 = $\frac{v}{4}$

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