滑块木板碰撞分析

创建于：2025年1月18日

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问题

如图所示，质量.M=0.5kg、半径R=0.2m的\frac {1}{4}光滑圆弧轨道 C 静置于光滑水平面上，O为圆心，P为轨道最低点，圆弧轨道C左侧地面上静置一质量m=0.3kg的水平木板B，木板B的上表面与圆弧轨道 P 点等高。某时刻质量m₀=1.2kg的滑块 A 以水平速度v₀=10m/s从木板B 左端滑上，当木板B 与圆弧轨道C碰撞前瞬间，滑块A 与木板B 恰好共速，且滑块A刚好位于木板B的右端。木板B 与圆弧轨道C碰后粘合在一起，木板B与滑块 A 之间的动摩擦因数μ=0.8,重力加速度g=10m/s²,滑块 A 可视为质点，求：
(1)木板B 的长度L;
(2)从滑块开始运动至木板与圆弧轨道碰后瞬间的过程中产生的热量Q；
(3)滑块 A 相对P点能够上升的最大高度h。

思考

Analyzing the problem

I'm looking at a frictionless quarter circle track and a block sliding on top of another block. The friction coefficient, masses, and initial velocity are specified.

Analyzing collision dynamics

First, A and B move together at 10 m/s to the track. After collision, they form one object of 0.8 kg. Calculating heat generated and A's height on the track next.

Tracking friction dynamics

I'm curious about how friction accelerates B while decelerating A, achieving a shared velocity. This involves calculating the force, distance, and frictional heat.

Evaluating horizontal forces

I'm zeroing in on the lack of external horizontal forces on A and B, with friction forces as internal interactions. This simplifies the system dynamics for further analysis.

Tracking momentum shifts

OK, I’m piecing together momentum changes for A and B, who ultimately share a 8 m/s velocity. This reveals the conserved momentum throughout their motion.

Understanding friction dynamics

OK, I’m thinking about how the 0.8 friction coefficient causes a 9.6 N force, accelerating B to the right and slowing A to the left as they slide relative to each other.

Figuring out displacement

I’m calculating relative displacement by noting the 9.6 N force affecting A and B’s velocities over 0.25 seconds, leading to a 2 m displacement for A toward the left.

Calculating displacements

I’ve figured out the displacements for A and B: 2.25 m and 1 m. The relative displacement is 1.25 m, matching the acceleration over the 0.25-second interval.

Confirming measurements

I’m thinking through the relative displacement between A and B, which totals 1.25 m. Checking the frictional heat calculation, I confirm it equals 12 J. This approach ensures accuracy.

Confirming collision energy

OK, let's confirm this: comparing initial and final kinetic energy shows 12 J lost, matching friction heat calculation. Now, we can tackle the inelastic collision scenario between B and the track.

Analyzing collision dynamics

I’m thinking through the inelastic collision of B with the track. Momentum conservation helps find their shared velocity. A, unaffected by the collision, retains its constant velocity.

Tracking momentum shifts

OK, let me see: B and the track merge at 3 m/s. A maintains its 8 m/s velocity, unaffected by the collision—suggesting no active braking or external forces.

Assessing velocity dynamics

I'm thinking A, if it stays in contact with B, might decelerate to B's velocity of 3 m/s due to friction. I’m also considering if A might continue sliding at 8 m/s.

Weighing friction forces

Interestingly enough, friction between A and B might cause A to slow to B's 3 m/s velocity in the collision's short span, given the 9.6 N friction force.

Mapping movement

I’m curious if block A will slide onto the frictionless track, with no horizontal force. The motion involves a 3 m/s velocity for B and the track.

Evaluating energy dynamics

I’m curious about using energy analysis to determine A's maximum height relative to P: the track and B's 3 m/s velocity creates an initial 5 m/s relative to A.

Tracking A's journey

I'm figuring out whether block A carries on at 8 m/s or gets a boost onto the track after B and the track collide, while maintaining the same horizontal velocity.

Evaluating momentum

I’m considering how block A maintains its 3 m/s velocity and gains upward momentum with a 3 m/s horizontal normal force from the frictionless track as it climbs.

Breaking down the dynamics

I'm digging deep into horizontal momentum and mechanical energy conservation to track block A's movement up the frictionless track, with an initial velocity from the collision between B and the track.

Handling momentum and energy

Tackling horizontal momentum and mechanical energy conservation for the moving block paired with a frictionless quarter circle track, aiming to find the block's peak height.

Piecing together energy dynamics

I'm exploring how the block's energy changes as it ascends with the wedge, considering potential energy and kinetic energy to identify the maximum achievable height.

Piecing together dynamics

I'm charting the block's relative velocity of 5 m/s towards the track and the impact of the normal force once it ascends the track, considering the combined mass of the wedge and track.

Calculating momentum shifts

I'm analyzing how the block's horizontal momentum changes with the angled wedge's increasing velocity. This insight uncovers the potential for further movement or height.

Pushing boundaries

I'm looking at how the block lifts with the wedge, preserving energy without friction, aiming to find the maximum potential energy and subsequent height.

Assessing the wedge

Considering if the wedge rotates on the frictionless plane, examining the impact of angular momentum from the block's normal force. This leads me to think about rotational dynamics.

Examining rotation dynamics

I'm thinking through the potential for the quarter circle track to rotate due to the normal force. This would influence both the track and the block's movements on the frictionless plane.

Ensuring track stability

OK, let me see. Maintaining the track's horizontal translation without rotation is simple. The normal force from the ground prevents net torque, so the wedge remains steady.

Analyzing motion dynamics

Ensuring the track's horizontal movement without rotation, I examine the block's potential and kinetic energies. This approach highlights the continuity of horizontal momentum and energy conservation.

Assessing kinetic energy

OK, let me see. The total initial kinetic energy of the system—consisting of the block and wedge—is 42 J, derived from their respective velocities and masses.

Tracking equilibrium states

I'm clarifying that both the wedge and block stabilize when they share the same velocity. The max height achieved is 0.4 m with energies and constraints balanced.

Figuring out height options

I’m thinking about the block's maximum height on the arc, considering the frictionless track, gravitational potential energy, and the possibility of the wedge recoiling.

Understanding velocity dynamics

I’m piecing together the block's velocity relative to the ground, considering the wedge and arc constraints. The block's tangential velocity aligns with the arc at a fixed radial distance.

Tracking horizontal dynamics

I’m thinking through how the block's horizontal position hinges on the wedge's velocity and the radial distance determining the block's tangential velocity.

Identifying the center

I’m figuring out the block's maximum potential energy by tweaking its vertical position, considering the wedge's recoil and the system's horizontal momentum constraints.

Tracking momentum and energy

I'm looking at the wedge-block system's horizontal momentum and mechanical energy. The block's velocity is purely tangential, and the radius stays fixed at 0.2.

Selecting angles

OK, I'm defining the angle from the vertical to simplify the problem. This approach efficiently identifies the maximum height, aligning with the quarter circle track's geometry.

Evaluating velocity impacts

I'm curious how the wedge's movement influences the block's horizontal and tangential velocities. The block's horizontal velocity stays constant due to the lack of horizontal forces.

Addressing horizontal acceleration

I’m mapping out how the wedge's normal force influences the block's horizontal velocity. This exploration leads me to conclude that the wedge-block system preserves consistent horizontal velocity.

Understanding momentum shifts

I’m piecing together how zero net external force for the wedge-block system translates to consistent horizontal and vertical velocities for the block, influenced by the wedge's movement.

Assessing force dynamics

I'm mapping out how the absence of friction results in the block's zero horizontal acceleration by analyzing the forces and the wedge's free movement on a frictionless plane.

Tracking horizontal movements

I'm thinking through how the wedge removes friction's horizontal impact, allowing only vertical motion for the block while the wedge slides laterally. This simplifies the problem significantly.

Keeping the momentum balance

I'm mapping out how the block's horizontal momentum at 9.6 m/s needs the wedge to move at 3 m/s, ensuring consistent energy and momentum conservation on the frictionless track.

Analyzing block's trajectory

Taking a closer look at the block's movement on a frictionless quarter circle track. The normal force is radial, affecting the block's tangential velocity relative to the wedge.

Tracking energy exchange

I'm focusing on how the block's mechanical energy relative to the wedge stays the same, while the wedge itself gains kinetic energy. This balances overall energy in the system.

Mapping energy patterns

Thinking through how the block's kinetic and potential energy stays stable relative to the wedge, while the wedge, as a non-inertial frame, influences the system's energy dynamics.

Analyzing mechanical energy

Tracking the block-wedge system's total mechanical energy, factoring in horizontal momentum, the block's vertical displacement, and kinetic energy for a comprehensive view.

Assessing mechanical energy

Taking a closer look at the block-wedge system's total mechanical energy, I factor in horizontal momentum, the block's vertical displacement, and kinetic energy to gauge the maximum height and velocity.

Piecing together energy movements

I'm working through total mechanical energy, checking horizontal momentum, the block's vertical rise, and internal kinetic energy to figure out the maximum height and velocity.

Charting shifts and forces

I'm curious about how the system's vertical center of mass changes with the block's height, while the wedge stays at zero vertical velocity on the frictionless plane.

Energy shifts

OK, let me see. If the block's vertical potential energy increases by 1.2gy, that has to come from internal kinetic energy or the center of mass's kinetic energy.

Balancing energy dynamics

I'm mapping out how the wedge's gravity and the lack of friction ensure the block's mechanical energy stays steady, with no work done on the wedge and internal shifts keeping the system intact.

Balancing energy shifts

I'm weighing the system's total mechanical energy, with the block's potential and internal kinetic energy, keeping in mind the center of mass's steady horizontal velocity.

Ensuring energy stability

With horizontal movement steady and internal energy shifts accounted for, structural stability and mechanical efficiency are ensured by the spillway design.

Tracking momentum gain

I'm mapping out the block's horizontal momentum, with the wedge accelerating, aiming for the block to reach the moment P where the wedge could start to lift.

Analyzing energy dynamics

I’m mapping out the twisting movement of energy through the system, factoring in frictionless constraints and gravitational potential, to pinpoint the block's peak height on the frictionless quarter-circle track.

Piecing together

Considering horizontal momentum and gravitational potential energy, assuming the wedge stays grounded and the maximum height is observed when the block reaches the quarter circle's peak.

Evaluating energy dynamics

The block's 38.4 J of initial kinetic energy is more than enough to overcome the 2.4 J potential energy change with the frictionless quarter circle track, ensuring it can reach the top.

Evaluating energy requirements

I’m thinking the block has sufficient energy to reach the top of the track. Ensuring the normal force remains positive, the block will stay on the quarter circle track.

Assessing energy balance

Ensuring the block remains in contact with the frictionless quarter circle track, while analyzing energy dynamics, is key to understanding its maximum height potential.

Assessing mechanical energy

I'm analyzing how mechanical energy is conserved with a frictionless track. The block's initial 5 m/s speed leads to a specific height estimate, factoring in radial normal force and projectile motion.

Mapping out mechanical energy

I'm assessing the block's mechanical energy with a fixed track: 5 m/s speed, radial normal force, and centripetal acceleration. This helps gauge the maximum height it can reach.

Balancing forces

Ensuring the block's radial normal force sustains centripetal acceleration to maintain contact. The maximum height is where the normal force yields zero, revealing the pivotal angle θ.

Identifying contact loss

Balancing mechanical energy and potential energy reveals the angle θ where the block loses contact with the track, ensuring the radial force balances gravity to maintain contact.

Evaluating contact conditions

I'm noting that the block doesn't lose contact with the track based on the failed real solution for losing contact as the normal force equals zero. The problem is reassuring.

Evaluating potential

OK, let me see. The block can ascend the frictionless quarter-circle track, given the failed real solution for losing contact as the normal force equals zero.

Tracking ascent dynamics

I'm curious if the block can reach the top of the frictionless quarter-circle track. The calculations show an initial speed of 4.58 m/s at 90 degrees, confirming it can.

Confirming the ascent

I'm ensuring that with ample energy, the block can climb to a height of 0.2 meters on the frictionless quarter-circle track. This aligns with the track's geometry.

Calculating and climbing

OK, here's the plan: the block's potential energy is 2.4 J, its initial kinetic energy is 15 J, and with 12.6 J left, it can definitely reach 0.2 m.

Thought about mechanics and energy conservation for 2m 35s

回答

下面给出一个比较详细的分析思路，供参考。题目所涉过程可分为三个阶段来讨论：

题目回顾与已知条件

四分之一圆弧轨道 $C$ ，质量 $M=0.5\,\mathrm{kg}$ ，半径 $R=0.2\,\mathrm{m}$ ，放在光滑水平面上，可自由平动（无摩擦，无外力束缚）。
木板 $B$ ，质量 $m=0.3\,\mathrm{kg}$ ，放在同一光滑平面上（对地无摩擦），上表面与轨道最低点 $P$ 齐平。
滑块 $A$ ，质量 $m_0=1.2\,\mathrm{kg}$ ，以初速度 $v_0=10\,\mathrm{m/s}$ 从木板左端滑上木板；滑块与木板之间的动摩擦因数 $\mu=0.8$ 。
已知重力加速度 $g=10\,\mathrm{m/s^2}$ 。
发生的过程：
- 滑块从木板左端滑到木板右端，同时由于摩擦作用，木板也被带动。
- 碰撞前瞬间，滑块与木板速度相同，且滑块正好到达木板右端。
- 随后木板与轨道发生碰撞并粘合在一起(完全非弹性碰撞)。
- 问题(3)要求：此后滑块沿圆弧轨道上滑时(轨道与滑块间光滑无摩擦)，它相对于圆弧最低点 $P$ 能升到的最大高度 $h$ 。

本题三个小问分别是：

木板 $B$ 的长度 $L$ ；
从滑块开始运动到木板与轨道碰前瞬间，滑块与木板间因摩擦所产生的热量 $Q$ ；
碰撞并粘合后，滑块继续运动到最大上升点时，相对 $P$ 的最大高度 $h$ 。

第一阶段：滑块在木板上滑动并带动木板运动

阶段初：

滑块 $A$ 初速度 $v_0=10\,\mathrm{m/s}$ ，质量 $m_0=1.2\,\mathrm{kg}$ ；
木板 $B$ 静止，质量 $m=0.3\,\mathrm{kg}$ 。

阶段末(碰撞前瞬间)：

题目给出：滑块与木板“恰好共速”，且滑块位于木板右端。

1. 利用动量守恒求它们最后的共同速度

这一阶段，滑块与木板之间只有接触摩擦，木板与地面光滑无摩擦，所以滑块+木板系统在水平方向上无外力，水平方向动量守恒。

初动量

p_\text{初} \;=\; m_0\,v_0 \;+\; m \cdot 0 \;=\; 1.2\times10 \;=\;12\,(\mathrm{kg\cdot m/s}).

末状态共同速度记为 $v_f$ ，总质量 $m_0 + m = 1.2 + 0.3 = 1.5\,\mathrm{kg}$ 。

(m_0 + m)\,v_f \;=\;12 \quad\Longrightarrow\quad v_f \;=\;\frac{12}{1.5} \;=\;8\,\mathrm{m/s}.

因此，在木板与轨道碰撞前一瞬间，滑块与木板的共同速度都是 $8\,\mathrm{m/s}$ 。

2. 求木板长度 $L$

滑块从左端滑到右端，相对于木板运动的那段“相对位移”即是木板的长度 $L$ 。可以先求出在这段时间内，滑块和木板各自在地面参考系中的位移，再做差。

设二者从相对速度开始到相对速度变为零(即二者同速)所用时间为 $t$ 。
摩擦力大小： $\displaystyle f = \mu\,m_0\,g =0.8\times1.2\times 10=9.6\,\mathrm{N}$ ，
方向：对滑块是阻力，对木板是正向拉力。

(1) 滑块的加速度

a_A = \frac{F_{\text{合}}}{m_0} = \frac{-\,9.6}{1.2} = -\,8\,\mathrm{m/s^2}.

滑块速度由 $10\,\mathrm{m/s}$ 减到 $8\,\mathrm{m/s}$ ，变化量 $\Delta v_A=-2\,\mathrm{m/s}$ ，
所需时间

t = \frac{\Delta v_A}{a_A} = \frac{-2}{-8} = 0.25\,\mathrm{s}.

(2) 木板的加速度

a_B = \frac{9.6}{0.3} = 32\,\mathrm{m/s^2}.

木板速度由 $0$ 增至 $8\,\mathrm{m/s}$ ，同样在时间 $t=0.25\,\mathrm{s}$ 内完成。

(3) 二者在地面系的位移

滑块位移

x_A = v_{A,\text{平均}} \,\times\, t = \frac{\,10 + 8\,}{2}\,\times\,0.25 = 9\times0.25 = 2.25\,\mathrm{m}.

木板位移

x_B = v_{B,\text{平均}} \,\times\, t = \frac{\,0 + 8\,}{2}\,\times\,0.25 = 4\times0.25 = 1.00\,\mathrm{m}.

(4) 滑块相对木板的位移

L \;=\; x_A - x_B \;=\;2.25 - 1.00 \;=\;1.25\,\mathrm{m}.

故木板的长度

\boxed{\,L=1.25\,\mathrm{m}\,}.

3. 求产生的热量 $Q$

这一阶段的能量损失全部是滑块与木板之间的摩擦生热。

也可以直接用系统总动能的变化量来求：
- 初始(滑块10,m/s,木板0,m/s)的总动能
$E_\text{初} = \tfrac12\,m_0\,v_0^2 = \tfrac12\times1.2\times 10^2 = 60\,\mathrm{J}.$
- 末态(滑块8,m/s,木板8,m/s)的总动能
$E_\text{末} = \tfrac12\,(m_0+m)\,(8^2) = \tfrac12\times1.5\times 64 = 48\,\mathrm{J}.$
- 摩擦所耗(转化为内能/热量)
$Q = E_\text{初}-E_\text{末} = 60 - 48 = 12\,\mathrm{J}.$

或者用“摩擦力 $\times$ 相对位移”也能得到同样结果：

Q = f \cdot (x_A - x_B) = 9.6\times 1.25=12\,\mathrm{J}.

因此产生的热量

\boxed{\,Q=12\,\mathrm{J}\,}.

第二阶段：木板与轨道的碰撞并粘合

在上述末瞬间，木板和滑块同速 $8\,\mathrm{m/s}$ ，轨道静止。题目说木板与轨道发生完全非弹性碰撞并黏在一起。

碰撞前：
- 木板质量 $m=0.3\,\mathrm{kg}$ ，速度 $8\,\mathrm{m/s}$ ；
- 轨道质量 $M=0.5\,\mathrm{kg}$ ，速度 $0$ .
碰撞后：
- 合体(记为“楔形体”或“轨道+木板”)质量 $m+M=0.8\,\mathrm{kg}$ 。
- 由动量守恒得合体速度 $V_{\text{wedge}}$ 。滑块 $A$ 此刻不参与这个碰撞冲量(题目已说明“碰撞前瞬间，滑块和木板共速”，但碰撞只发生在木板和轨道之间)。

碰撞前合体(仅木板部分)的动量：

p_\text{前} \;=\;0.3\times8 \;=\;2.4\,(\mathrm{kg\cdot m/s}).

碰撞后合体质量 $0.3+0.5=0.8$ ，故

0.8\,V_{\text{wedge}} \;=\;2.4 \quad\Longrightarrow\quad V_{\text{wedge}}\;=\;3\,\mathrm{m/s}.

与此同时，滑块仍然以 $8\,\mathrm{m/s}$ 的速度向右运动(碰撞是木板与轨道的粘合，滑块并未受到该瞬间的碰撞冲击)。因此：

碰撞后，轨道+木板整体速度为 $3\,\mathrm{m/s}$ ；
滑块速度仍是 $8\,\mathrm{m/s}$ 。

由此可见，碰撞后的那一刹那，滑块相对于“轨道+木板”有 $8-3=5\,\mathrm{m/s}$ 的相对速度(向右)。

第三阶段：滑块沿光滑四分之一圆弧上滑的最大高度

从碰撞后一瞬间开始：

“轨道+木板” 合体(以下简称“楔块”)质量 $0.8\,\mathrm{kg}$ ，水平速度初值 $3\,\mathrm{m/s}$ ；
滑块质量 $1.2\,\mathrm{kg}$ ，水平速度初值 $8\,\mathrm{m/s}$ ；
轨道与滑块无摩擦，且轨道可在光滑平面上自由平动(仅水平方向)，不考虑其转动(通常认为轨道底面支撑在光滑平面上，支撑反力可偏移以抵消转动趋势)。

1. 是否需要担心“系统因为相互作用而损失能量”？

在这个阶段：

滑块与轨道间无摩擦，只有法向约束力，法向力属于内力，不会产生能量耗散；
轨道与地面也无摩擦，不存在外力耗散。

所以“滑块+楔块”系统的机械能守恒(重力是保守力，地面对楔块的正压力不做水平功，也不做竖直功——楔块并未离开地面)。同时，水平方向也无外力，故总水平动量守恒。

2. 是否可能“滑块在轨道上脱离接触”？

要判断滑块能否一直紧贴轨道上滑，需要看它在圆弧上的法向加速度是否能由正常力+重力的分量来提供。若某处所需向心力大于重力的相关分量，就可能失去接触；或者等效地，用“轨道对滑块的法向力不能为负”做判断。

后面计算会发现：滑块初始相对轨道速度就已经较大(5m/s)，上升高度也仅 $R=0.2\,\mathrm{m}$ ，所需能量并不大，事实上它不但不会中途脱离，还能带着多余速度到达这段轨道顶端。

3. 直接判断能否到达四分之一圆弧的最顶端

半径 $R=0.2\,\mathrm{m}$ ，四分之一圆弧的最高点相对最低点 $P$ 的高差是 $0.2\,\mathrm{m}$ 。
问题只要求**“相对 $P$ 能够上升的最大高度”**，且轨道只有这一段弧(从最低点到 $90^\circ$ 处)。只要滑块有足够能量，就能到达弧的最上端(高度 $0.2\,\mathrm{m}$ )。

一个常用而简化的判据是：把“轨道视为静止”这一参考系看成“虚拟固定参考系”，则滑块进入圆弧时的相对速度为

v_{\text{相对}} = 8 - 3 = 5\,\mathrm{m/s}.

若轨道真是固定不动，无摩擦，滑块以水平初速 $5\,\mathrm{m/s}$ 沿半径 $R=0.2$ 的圆弧上滑，其势能增量为

\Delta U = m_0\,g\,\Delta h = 1.2 \times 10 \times 0.2 = 2.4\,\mathrm{J}.

而初始动能(在这“轨道静止”想象系中)是

E_{\text{初}} = \tfrac12\,m_0\,\bigl(5^2\bigr) = 0.5\times1.2\times25 = 15\,\mathrm{J}.

显然 $15\,\mathrm{J}$ 远大于上升所需的 $2.4\,\mathrm{J}$ ，滑块还有剩余的动能；因此它不仅能到达 $0.2\,\mathrm{m}$ 的高度，而且到达顶部时速度仍不为零，不会中途停下或脱轨。

由此可见，它必然能爬到该四分之一圆弧的最高点。因此，上升的最大高度即为整段弧的高度：

\boxed{\,h = 0.20\,\mathrm{m}\,}.

小结与答案

木板长度

L \;=\;1.25\,\mathrm{m}.

第一阶段(滑块与木板相对滑动)所产生的总热量

Q = 12\,\mathrm{J}.

滑块相对 $P$ 能上升的最大高度

h = 0.20\,\mathrm{m}.

问题

思考

回答