Question

По горизонтальному столу скользит брусок и сталкивается с покоящимся бруском такой же массы, имея перед ударом скорость 120 см/с. Удар абсолютно упругий. Все скорости направлены вдоль одной прямой. Коэффициент трения брусков о стол равен 0,15. Найдите расстояние (в см) между брусками после их остановки. Принять g=10 м/с 2 . Если ответ не целый, округлите его до сотых.

Answer 1

При данном абсолютно упругом ударе первый брусок остановится, а второй приобретет импульс первого, то есть начнет движение с нач. скоростью v0 = 1,2 м/c. Его движение будет равнозамедленным с ускорением, равным по модулю kg.

Из формул кинематики пройденный путь:

S = (v0^2)/(2kg) = 0,48 m

Ответ: 0,48 м

Answer 2

Problem Analysis

We are given a horizontal table on which a block is sliding and collides with another stationary block of the same mass. The sliding block has an initial velocity of 120 cm/s before the collision. The collision is perfectly elastic, and the coefficients of friction between the blocks and the table are known. We need to find the distance between the blocks after they come to a stop.

Solution

To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.

1. Conservation of Momentum: According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the second block is stationary, its initial momentum is zero. Therefore, the momentum of the sliding block before the collision is equal to the momentum of both blocks after the collision.

2. Conservation of Kinetic Energy: In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. We can use this principle to find the final velocities of the blocks after the collision.

Let's calculate the final velocities of the blocks and then find the distance between them.

Calculation

Given: - Initial velocity of the sliding block (v1) = 120 cm/s - Coefficient of friction between the blocks and the table (μ) = 0.15 - Acceleration due to gravity (g) = 10 m/s^2

First, let's convert the initial velocity to m/s: v1 = 120 cm/s = 1.2 m/s

Next, let's calculate the final velocities of the blocks after the collision using the principles of conservation of momentum and conservation of kinetic energy.

Let: - Mass of each block (m) = m1 = m2 (since both blocks have the same mass)

According to the conservation of momentum: m1 * v1 = (m1 + m2) * v1f v1f = (m1 * v1) / (m1 + m2) ---(1)

According to the conservation of kinetic energy: (1/2) * m1 * v1^2 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2 v2f = sqrt((m1 * v1^2 - m1 * v1f^2) / m2) ---(2)

Now, let's calculate the final velocities of the blocks: Substituting the values in equations (1) and (2): v1f = (m1 * v1) / (m1 + m2) v2f = sqrt((m1 * v1^2 - m1 * v1f^2) / m2)

Next, let's calculate the distance between the blocks after they come to a stop. We can use the equation of motion to calculate the distance traveled by the blocks.

Let: - Distance between the blocks after they come to a stop (d) - Time taken for the blocks to come to a stop (t)

Using the equation of motion: d = v1f * t + (1/2) * (-μ * g) * t^2 ---(3)

Since the blocks come to a stop, the final velocities of both blocks are zero. Therefore, v1f = v2f = 0.

Substituting v1f = 0 in equation (3): d = (1/2) * (-μ * g) * t^2

To find the time taken for the blocks to come to a stop, we can use the equation of motion for the sliding block: v1f = v1 + (-μ * g) * t 0 = v1 + (-μ * g) * t

Solving for t: t = -v1 / (μ * g)

Now, let's substitute the values and calculate the distance between the blocks.

Calculation:

Given: - v1 = 1.2 m/s - μ = 0.15 - g = 10 m/s^2

Calculating t: t = -v1 / (μ * g) = -1.2 / (0.15 * 10) = -1.2 / 1.5 = -0.8 s

Since time cannot be negative, we take the absolute value of t: t = 0.8 s

Calculating d: d = (1/2) * (-μ * g) * t^2 = (1/2) * (-0.15 * 10) * (0.8)^2 = -0.6 * 0.15 * 0.64 = -0.0576 m

Since distance cannot be negative, we take the absolute value of d: d = 0.0576 m

Finally, let's convert the distance to centimeters and round it to two decimal places: d = 0.0576 m * 100 cm/m = 5.76 cm

Answer

The distance between the blocks after they come to a stop is 5.76 cm.

Note: The negative sign in the calculation of distance indicates that the blocks move in opposite directions after the collision. However, we take the absolute value of the distance to represent the magnitude of the separation between the blocks.