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Momentum
Definition:Is the product of body mass in the speed at that moment.K = k × pWhere:K: is the amount of movement.K: is the mass of the body.P: is the speed of the body.
Momentum is the amount of bound, and have the same direction is the direction of speed, so you must take into account when calculating the reference speed momentum.
A unit measuring the amount of move:Unit momentum = mass of unit mass and unit speed ×Unit amount of movement = kg × m / s
Example:Find the amount of momentum the ability of international unity in the following cases:A - k = 28 g, p = 100 cm / sK = k × pK = 0.028 × 1K = .028 kg. M / s
B - k = 2 tons, 60 km / hK = k × pK = 2000 × 16.67K = 33 340 kg. M / s
C - k = 5 kg, p = 30 m / sK = k × pK = 5 × 30K = 150 kg. M / s
Example:2 kg body mass fell from a height of 160 cm from the surface of the earth, calculate the amount of body movement derived by the moment of arrival to the surface of the earth.Solution
K = 2 kgP =?
Create the speed of the body the moment he arrived the Earth's surface:P 2 = p. 2 + 2 c PP 2 = 0 + 2 × 10 × 1.6P 2 = 32P = 5.66 m / s
K = k × pK = 2 × 5.66K = 11.32 kg. M / s
Example:If the amount of movement of the body mass of 14 kg at a given moment is 84 kg. Meters / sec. What is the speed of the body at that moment?K = k × p84 = 14 × pP = 6 m / s
Change in the amount of movement:
Suppose a body mass (k) affected the power of (s) changed the speed of the p. To G is:The change in momentum = final momentum - momentum primaryK = k p - k p.K = k (p - p.)
Note:Must take into account the signal of p, p.If p. , P in the same direction have the same signal.If they were in opposite directions, the Aharthma be different.And preferably at different directional two-speed signal as p is positive, the signal p. Is negative, and the direction of change in the amount of movement is the direction of p.It is estimated that the change in the amount of movement in the same unit measuring the amount of movement.
Example:Body mass 180 g moving in a straight line, if the changed speed of 9 m / s to 63 km / h in the same direction. Calculate the change in the amount of movement.SolutionK = 0.18 kgP. = 9 m / sP = 17.5 m / s (from about km to meters, hours to seconds)
K = k (p - p.)K = 0.18 (17.5 - 9)K = 1.53 kg. M / s
Example:Thrown ball mass 0.125 kg vertically to higher speeds of 8 m / s, calculate the change in the amount of movement when the height of the 2 m from the point of ejaculation is emerging.K = 0.125 kgP. = 8 m / sP =?
Create speed of the ball when you are on the rise 2 meters from the point of ejaculation:P 2 = p. 2 + 2 c PP 2 = 64 + 2 × (- 10) × 2P 2 = 24P = 4.89 m / s
K = k (p - p.)K = 0.125 (4.89 - 8)K = - 0.389 kg. M / sWhat is a negative reference?
Rate of change in the amount of move:
Reflects the amount of change in the amount of movement per unit time as the rate of change in the amount of movement.
Rate of change in the amount of movement = k p / nRate of change in the amount of movement = k × cRate of change in the amount of movement = s
Where:C = p / n
The unit of measurement the rate of change in the amount of movement:Is a unit of measure of force, which NewtonNewton is equivalent to kg × m / s 2
Example:Railcars 2 ton mass moving at 9 km / hour. Crashed into a barrier at the end of the line bounced back at 1.5 m / s. Answer the following:A) Calculate the change in the amount of movement as a result of this collision.B) If this happens regularly change in a time of 0.5 seconds founding rate of change in the amount of movement of the vehicle.SolutionK = 2000 kg (about one ton to kg)P. = - 2.5 m / s (a negative signal, and convert from kilometers to meters, hours to seconds)P = 1.5 m / s
A - change in momentum:K = k (p - p.)K = 2000 (1.5 - (- 2.5))K = 8000 kg. M / s
B - rate of change in the amount of move:K / time = 8000 / 0.5K / time = 16,000 Nm
Example:Hit a ball of rubber mass 80 grams of 360 cm height above the ground, bounced and hit the the top of the vertical distance of 130 cm. Answer the following:A) Calculate the change in the amount of movement of the ball as a result of a collision with the ground.B) If this happens regularly at the changing time of 0.1 seconds, Vahsb rate of change in the amount of movement as a result of this collision.C) How much regression coefficient? - We will discuss it in some detail later -Solution
K = 0.08 kgF (fall) = 3.6 metersP (bounce) = 1.3 meters
Create speed of the ball the moment of collision with the surface of the Earth:P 2 = p. 2 + 2 c PP 2 = 0 + 2 × 10 × 3.6P 2 = 72P = 8.5 m / sP. = - 8.5 m / s
Create speed of the ball rebounded from the moment of the Earth's surface:P 2 = p. 2 + 2 c P0 = p. 2 + 2 × (- 10) × 1.3P = 26 0.2P. = 5.1 m / sP = 5.1 m / s
A - change in momentum:K = k (p - p.)K = 0.08 (5.1 - (- 8.5))K = 1.088 kg. M / s
B - rate of change in the amount of move:K / time = 1.088 / 0.1K / time = 10.88 Newton
C - regression coefficient (t)T = - (p 1 - p 2) / (p 1 - p 2)T = - (5.1 - 0) / (-8.5 -0)T = 0.6
Definition:Is the product of body mass in the speed at that moment.K = k × pWhere:K: is the amount of movement.K: is the mass of the body.P: is the speed of the body.
Momentum is the amount of bound, and have the same direction is the direction of speed, so you must take into account when calculating the reference speed momentum.
A unit measuring the amount of move:Unit momentum = mass of unit mass and unit speed ×Unit amount of movement = kg × m / s
Example:Find the amount of momentum the ability of international unity in the following cases:A - k = 28 g, p = 100 cm / sK = k × pK = 0.028 × 1K = .028 kg. M / s
B - k = 2 tons, 60 km / hK = k × pK = 2000 × 16.67K = 33 340 kg. M / s
C - k = 5 kg, p = 30 m / sK = k × pK = 5 × 30K = 150 kg. M / s
Example:2 kg body mass fell from a height of 160 cm from the surface of the earth, calculate the amount of body movement derived by the moment of arrival to the surface of the earth.Solution
K = 2 kgP =?
Create the speed of the body the moment he arrived the Earth's surface:P 2 = p. 2 + 2 c PP 2 = 0 + 2 × 10 × 1.6P 2 = 32P = 5.66 m / s
K = k × pK = 2 × 5.66K = 11.32 kg. M / s
Example:If the amount of movement of the body mass of 14 kg at a given moment is 84 kg. Meters / sec. What is the speed of the body at that moment?K = k × p84 = 14 × pP = 6 m / s
Change in the amount of movement:
Suppose a body mass (k) affected the power of (s) changed the speed of the p. To G is:The change in momentum = final momentum - momentum primaryK = k p - k p.K = k (p - p.)
Note:Must take into account the signal of p, p.If p. , P in the same direction have the same signal.If they were in opposite directions, the Aharthma be different.And preferably at different directional two-speed signal as p is positive, the signal p. Is negative, and the direction of change in the amount of movement is the direction of p.It is estimated that the change in the amount of movement in the same unit measuring the amount of movement.
Example:Body mass 180 g moving in a straight line, if the changed speed of 9 m / s to 63 km / h in the same direction. Calculate the change in the amount of movement.SolutionK = 0.18 kgP. = 9 m / sP = 17.5 m / s (from about km to meters, hours to seconds)
K = k (p - p.)K = 0.18 (17.5 - 9)K = 1.53 kg. M / s
Example:Thrown ball mass 0.125 kg vertically to higher speeds of 8 m / s, calculate the change in the amount of movement when the height of the 2 m from the point of ejaculation is emerging.K = 0.125 kgP. = 8 m / sP =?
Create speed of the ball when you are on the rise 2 meters from the point of ejaculation:P 2 = p. 2 + 2 c PP 2 = 64 + 2 × (- 10) × 2P 2 = 24P = 4.89 m / s
K = k (p - p.)K = 0.125 (4.89 - 8)K = - 0.389 kg. M / sWhat is a negative reference?
Rate of change in the amount of move:
Reflects the amount of change in the amount of movement per unit time as the rate of change in the amount of movement.
Rate of change in the amount of movement = k p / nRate of change in the amount of movement = k × cRate of change in the amount of movement = s
Where:C = p / n
The unit of measurement the rate of change in the amount of movement:Is a unit of measure of force, which NewtonNewton is equivalent to kg × m / s 2
Example:Railcars 2 ton mass moving at 9 km / hour. Crashed into a barrier at the end of the line bounced back at 1.5 m / s. Answer the following:A) Calculate the change in the amount of movement as a result of this collision.B) If this happens regularly change in a time of 0.5 seconds founding rate of change in the amount of movement of the vehicle.SolutionK = 2000 kg (about one ton to kg)P. = - 2.5 m / s (a negative signal, and convert from kilometers to meters, hours to seconds)P = 1.5 m / s
A - change in momentum:K = k (p - p.)K = 2000 (1.5 - (- 2.5))K = 8000 kg. M / s
B - rate of change in the amount of move:K / time = 8000 / 0.5K / time = 16,000 Nm
Example:Hit a ball of rubber mass 80 grams of 360 cm height above the ground, bounced and hit the the top of the vertical distance of 130 cm. Answer the following:A) Calculate the change in the amount of movement of the ball as a result of a collision with the ground.B) If this happens regularly at the changing time of 0.1 seconds, Vahsb rate of change in the amount of movement as a result of this collision.C) How much regression coefficient? - We will discuss it in some detail later -Solution
K = 0.08 kgF (fall) = 3.6 metersP (bounce) = 1.3 meters
Create speed of the ball the moment of collision with the surface of the Earth:P 2 = p. 2 + 2 c PP 2 = 0 + 2 × 10 × 3.6P 2 = 72P = 8.5 m / sP. = - 8.5 m / s
Create speed of the ball rebounded from the moment of the Earth's surface:P 2 = p. 2 + 2 c P0 = p. 2 + 2 × (- 10) × 1.3P = 26 0.2P. = 5.1 m / sP = 5.1 m / s
A - change in momentum:K = k (p - p.)K = 0.08 (5.1 - (- 8.5))K = 1.088 kg. M / s
B - rate of change in the amount of move:K / time = 1.088 / 0.1K / time = 10.88 Newton
C - regression coefficient (t)T = - (p 1 - p 2) / (p 1 - p 2)T = - (5.1 - 0) / (-8.5 -0)T = 0.6
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