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klemol [59]
3 years ago
10

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

Physics
1 answer:
madam [21]3 years ago
3 0

Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be 0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}. Accordingly, the ball's momentum at that moment would be p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right).

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2 years ago
A perfectly elastic collision occurs between a 15.0-kg mass traveling at 3.50 m/s and a 9.00-kg mass traveling at 2.35 m/s. if t
BaLLatris [955]
Momentum is conserved in a collision. Momentum is mass*velocity, so you can find your answer by calculating initial and final momentums and setting them equal to each other.

15kg * 3.50 m/s + 9kg * 2.35 m/s = 73.65 kg m/s

73.65 = 9 * 2.8 + 15x

solve for x
x= 3.23

The final velocity is 3.23 m/s
5 0
3 years ago
A cord is attached to the box and run through a pulley directly above the box, so that the cord is vertical. The free end of the
Harman [31]

Answer:

The answer is given here would be a simplified equation, seeing as there are some missing variables in the question.

<u>F1 = T- 46, 674.656 gm/s² </u>

Explanation:

<em>Note: Once we have the mass of the second object and/or acceleration of the cord, we can solve for the force of the ground acting on the box.</em>

To calculate the force caused by gravity on the basic pulley system we use the following equation:

F2 = M2 x g; where g= gravitational acceleration (a constant equal to 9.8 m/s²). The mass M2 = 10.5 lb = 4762.72g

∴ F2 = 4762.72g x 9.8 m/s²

= 46, 674.656 gm/s² or 46, 674.656 N

But since this F2 is acting in a downlowrd direction, it would be negative.

Tension of the cord, T = Mass, x × acceleration. ( x is in the pulley diagram)

⇒ F1 = T - F2

<u>F1 = T- 46, 674.656 gm/s² </u>

4 0
3 years ago
) A striker can give the ball an initial speed of 30m/s. Within what two elevation angles must he kick
sveticcg [70]

Answer:

  about 19.6° and 73.2°

Explanation:

The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...

  y = -4.9(x/s·sec(α))² +x·tan(α)

where s is the launch speed in meters per second.

We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):

  -13.6111·tan(α)² +50·tan(α) -16.0511 = 0

This has solutions ...

  tan(α) = 0.355408 or 3.31806

The corresponding angles are ...

  α = 19.5656° or 73.2282°

The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.

_____

I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.

5 0
3 years ago
A chevrolet corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. what
pychu [463]
By definition we have that the final speed is:
 Vf² = Vo² + 2 * a * d
 Where,
 Vo: Final speed
 a: acceleration
 d: distance.
 We cleared this expression the acceleration:
 a = (Vf²-Vo²) / (2 * d)
 Substituting the values:
 a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
 a = -77268 mi / h ^ 2
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
 First you must make a free body diagram and see the acceleration of the car:
 g = 32.2 feet / sec ^ 2
 a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
 a = -31.48 feet / sec ^ 2
 A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
 A = -22.07 feet / sec ^ 2
 Clearing the braking distance:
 Vf² = Vo² + 2 * a * d
 d = (Vf²-Vo²) / (2 * a)
 Substituting the values:
 d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
 d = 175.44 feet
 answer:
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
5 0
3 years ago
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