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As shown in the figure, a flat car with mass m moves in a straight line at speed V on a smooth horizontal plane. Today, an object with mass m is gently placed at the front end of the car. When the object is placed on the car, the horizontal speed relative to the ground is zero. Let the dynamic friction coefficient between the object and the car be μ， In order to prevent objects from sliding down from the trolley, the minimum length of the trolley is __
According to the conservation of momentum: MV = (M + m) V ′
According to the functional relationship: μ mgL=1
2mv2-1
2（M+m）v′2
Lianlide: l = MV2
two μ g(m+M)
So the answer is: MV2
two μ g(m+M)．
There are two trolleys with mass m of a and B, which move towards each other on the same straight line at the same speed V0 on a smooth horizontal plane. Car a moves to the right and car B moves to the left. There is a person with mass m on car a and no one on car B. ask the person on car a at least what horizontal speed to jump from car a to car B in order to avoid collision between the two cars. The process needs to be detailed
Take the right direction as the positive direction, take a as the research object (M + m) V0 = MV1 + mvav1 = [(M + m) V0 MVA] / M take B as the research object - MV0 + MVA = (M + m) v2v2 = (- MV0 + MVA) / (M + m) if the two vehicles do not collide, V1 ＜ = V2, get VA > = (2m ^ 2 + 2mm + m ^ 2) / (m ^ 2 + 2mm), so the minimum speed of VA is (2m ^ 2 + 2mm + m ^ 2) / (m ^ 2
The trolley with mass m is stationary on a smooth horizontal plane, and the person with mass m stands at the left end of the trolley. In the process of walking from the left end to the right end of the trolley () A. If a person suddenly stops relative to the car while walking, the speed of the car relative to the ground will turn to the right B. The greater the average speed of people walking on the car, the greater the distance the car moves on the ground when walking to the right end C. The smaller the average speed of people walking on the car, the greater the distance the car moves on the ground when walking to the right end D. No matter what average speed a person walks at, the distance a car moves on the ground is the same
The system composed of human and vehicle is selected as the research object, and the direction of human walking is the positive direction; A. In the process of walking on the car, the momentum in the horizontal direction is conserved. Because the momentum of the system composed of man and car is conserved in the horizontal direction, if the person suddenly stops relative to the car during walking, the speed of the car relative to the ground is 0
The wooden board B with a mass of M = 3kg is placed on a smooth horizontal plane, and a wooden block a with a mass of M = 2kg rushes to the left end of the wooden board at a speed of V0 = 5m / s. when AB is relatively stationary, the wooden board just hits the fixed wall C, the collision time between B and C is very short and there is no energy loss, and then the wooden block a just stops at the right end of wooden board B. It is known that the dynamic friction factor between a and B is u = 0.6, g = 10m / S2. Find: (1) What is the speed of the final boards and blocks? (2) What is the length L of the board?
When AB is relatively stationary, the speed is v '; The speed of the final board and block is v
mv0=(M+m)v'
v'=2m/s
mgus=1/2mv0^2-1/2(M+m)v'^2
s=5/4m
Mv'=(M+m)v
V = 6 / 5m / s, opposite direction
Lmgu=1/2mv0^2-1/2(M+m)v^2
L=1.71m
Throw the object with mass m = 0.5m/s horizontally at the initial speed V0 = 6m / s. After 0.8s, how much does the momentum of the object increase (g takes 10m / S2)
The momentum increase of an object is caused by the action of gravity, that is, the impulse of gravity
Δ p=Gt=mgt=0.5 × ten × 0.8kg·m/s=4kg·m/s
Note: momentum is a vector, and the change of momentum is not the change of momentum
As shown in the figure, the object with mass m (which can be regarded as a particle) slides onto the trolley with mass m which was originally stationary on a smooth horizontal plane at horizontal speed v0, The dynamic friction coefficient between the object and the trolley surface is u, and the trolley is long enough to calculate the displacement of the trolley from the time when the object is drawn on the trolley to the time when the trolley is relatively stationary
Conservation of momentum:
mv0=(M+m)v
Speed of trolley v = MV0 / (M + m)
Friction = UMG
Acceleration of trolley = UG
2ug*S=v^2
Displacement of trolley passing s = m ² v0 ²/ [(M+m) ² 2ug]
RELATED INFORMATIONS
- 1. The object with mass m slides to the trolley with mass m originally stationary on the horizontal smooth plane at the horizontal initial speed V0, and the dynamic friction between the object and the trolley surface is μ, The trolley is long enough to calculate the displacement of the trolley from the time when the object slides onto the trolley to the time when it is relatively stationary with the trolley
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