A person weighing 80 kg moves from bow to stern in a stationary boat of length s = 5 m. What is the mass of the boat if during this transition it has moved L = 2 m in still water? Ignore water resistance. v1. v2. 1. O. X. L. 0 =. m1v1. + (m1 + m2)v2. 2. V =s/t. - m1v1. 3.0=. + (m1 + m2)v2. 0 =. - m1s|t. + (m1 + m2)L|t. M1s|l – m1 = 80 kg*5 m/ 2 m – 80 kg = 120 kg. 4. m2 =.
Slide 10 from the presentation "Problems for maintaining momentum". The size of the archive with the presentation is 227 KB.Physics 9th grade
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1. a) Can the total momentum of two bodies be less than the momentum of one of these bodies? Explain your answer.
b) A skater who threw a stone weighing 2 kg horizontally at a speed of 15 m/s rolled back 62.5 cm. Determine the mass of the skater if the coefficient of friction of the skates on the ice is 0.02.
2. a) Why is it that a bullet fired from a gun cannot open the door, but makes a hole in it, whereas with the pressure of a finger it is easy to open the door, but it is impossible to make a hole.
b) A person weighing 60 kg moves from the bow to the stern of the boat. How far will a boat 3 m long move if its mass is 120 kg?
3. a) Two material points equal mass moving towards each other with equal absolute speeds. What is the total momentum of the points?
b) A person with a mass of 80 kg moves from bow to stern in a boat 5 m long. What is the mass of the boat if during this transition it moved in still water in the opposite direction by 2 m?
4. a) What should a person do to avoid falling through thin ice: run on the ice or stand on it?
b) A rope is pulled from the boat and fed to the longboat. The distance between them is 55 m. Determine the paths traveled by the boat and the longboat before they met. The weight of the boat is 300 kg, the weight of the longboat is 1200 kg. Neglect water resistance.
5. a) A flying bullet does not break the window glass, but creates a round hole in it. Why?
b) A skater weighing 70 kg, standing on ice, throws a stone weighing 3 kg in a horizontal direction at a speed of 8 m/s relative to the ice. Find the distance the skater will roll if the coefficient of friction is 0.02.
6. a) Could the hero of the book by E. Raspe, Baron Munchausen, according to his story, actually pull himself and his horse out of the swamp?
b) The projectile breaks into two identical parts at the top point of the trajectory at a height of 20 m. 1 second after the explosion, one part falls to the ground under the place where the explosion occurred. At what distance from the place of the shot will the second part of the projectile fall if the first part fell at a distance of 1000 m? Do not take into account the force of air resistance when solving the problem.
Solution.
We will connect the reference system with the surface of the Earth and consider it inertial. Axis OX direct horizontally, axis OY – vertically up.
Let us assume that the physical system includes only a boat and a person. Earth, air and water are external bodies in relation to the selected physical system.
The interaction of the system with them can be described using the corresponding forces. We can distinguish two states of the system: the beginning of the jump and the end of the jump. Even without taking into account the interaction with air, the physical “man-boat” system is not closed, because at the moment of jumping, the force of gravity directed vertically downwards acts on a person. Therefore, the total momentum vector of this system is not conserved, i.e. p 1 ≠ p 2 . However, in this case, the projection of the total momentum vector onto the horizontal direction (axis OX ), since external forces do not act in this direction (at the moment of the jump, the water resistance force is zero, since the boat is at rest).
The momentum vectors of the bodies of the system are shown in the figure.
Let us write down the conservation law for the horizontal component of momentum.
If we project vector quantities onto the axis OX
How can we find the speed of the boat after a person jumps?
To determine the distance the boat will move after the jump, consider the physical system “boat after the jump.”
The selected physical system is not closed, since it interacts with material objects that are not included in it. If you do not take into account the interaction of the boat with the air, then it is acted upon by: gravity m 1 g , caused by interaction with the gravitational field of the Earth; resistance force F with and buoyant force F in caused by interaction with water. Any open system can be described by the laws of kinematics, dynamics and the theorem on the change in kinetic energy.
We use the laws of kinematics and dynamics. The forces acting on the boat while moving are constant, so it will move in a straight line with constant acceleration. Thus,
The forces acting on the boat and the kinetic quantities characterizing its movement are shown in the figure on the left.
Let's take the origin of coordinates at the point on the surface of the water where the boat is at the moment of the jump, the axis OX we will direct the movement of the boat, the axis OY – vertically up. With this choice of coordinate system, the initial coordinate of the boat is zero, and the final coordinate l.
Therefore, if we project vector quantities onto the coordinate axes, taking into account the fact that the final speed of the boat v = 0, we get a system
Where
If we substitute the value into the last formula v 1 .
Physics problem - 1772
2017-01-04
The boat is motionless in still water. The person in the boat moves from the bow to the stern. How far will the boat move if the mass of the person is $m = 60 kg$, the mass of the boat is $M = 120 kg$, and the length of the boat is $l = 3 m$? Ignore water resistance.
Solution:
Let a person move from bow to stern uniformly over time $t$ (Fig.). Since we assumed that there are no external forces, the impulse of the boat-person system should not change, i.e., during the entire time the person is moving, the boat should move in the opposite direction with such a speed that the total impulse is equal to zero. Let the boat move in the opposite direction by a distance $x$ during the same time $t$. Then the speed of the person relative to the ground during this time was $(l - x)/t$, and the speed of the boat was $x/t$. The law of conservation of momentum gives
$m(l-x)/t - Mx/t=0$,
$x = ml/(M + m) = 1 m$.
The same result can be obtained based on the corollary arising from the law of conservation of momentum: in the absence of external forces, the center of mass of the system cannot move. When a person stands on the bow H of the boat, the center of mass of the boat - person system is on the vertical passing through point A, with CA = 0.5 m. When the person moves to the stern K, then the center of mass of the same system is on the vertical passing through point B, with BC = 0.5 m. Since during the person’s transition from bow to stern no external forces acted on the boat-man system, the center of mass of the system cannot move. To do this, the boat must move so that point B coincides with the previous position of point A, that is, the boat must move to the right by a distance BA equal to 1 m.