Question

Летящая горизонтально пуля попадает в лежащий на гладком столе деревянный брусок и застревает в нем. Масса пули m=10г, масса бруска М=100г, а скорость пули 110м/с. Определите кинетическую энергию бруска с пулей после удара. Сораняется ли полная механическая энергия в этом случае?

Answer 1

Воспользуемся законом сохранения к-ва движения -  внешних сил по оси х нет.

mv1=(M+m)v2→10*110=110*v2→ v2=10 м/с скорость бруска. 
кинетическая энергия бруска с пулей (M+m)v2²/2=0.11 кг*100 /2 м/с =
= 11/2=5,5 дж  
 
начальная кинетическая энергия пули mv1²/2=0.01*110²/2=60.5 дж
полная механическая энергия не сохраняется - ее энергии пули уходит на нагревание бруска

Answer 2

Дано    m=0,01 кг  M=0,1 кг    V=110 м/с   E- ?

по закону сохранения импульса   m*V=(m+M)*U

U=m*V/(m+M)=0,01*110/0,11=10 м/с
Е=(m+M)*U^2/=0,11*100/2=5,5 Дж

E=m*V^2/2=0,01*110^2/2=121/2=60,5 Дж >5,5 Дж
значит механическая энергия не сохраняется 

Answer 3

Problem Analysis

We are given the following information: - Mass of the bullet, m = 10 g - Mass of the wooden block, M = 100 g - Velocity of the bullet, v = 110 m/s

We need to determine the kinetic energy of the wooden block with the bullet after the impact and whether the total mechanical energy is conserved in this case.

Solution

To find the kinetic energy of the wooden block with the bullet after the impact, we can use the formula for kinetic energy:

Kinetic Energy (KE) = (1/2) * mass * velocity^2

Let's calculate the kinetic energy of the bullet first using the given mass and velocity:

KE_bullet = (1/2) * m * v^2

Substituting the values, we get:

KE_bullet = (1/2) * 10 g * (110 m/s)^2

To calculate the kinetic energy of the wooden block with the bullet after the impact, we need to consider the conservation of momentum. Since the bullet gets embedded in the wooden block, the final velocity of the combined system will be the same as the initial velocity of the bullet.

Let's calculate the velocity of the combined system after the impact:

m * v_bullet_initial = (m + M) * v_combined_final

Substituting the values, we get:

10 g * 110 m/s = (10 g + 100 g) * v_combined_final

Simplifying the equation, we find:

v_combined_final = (10 g * 110 m/s) / (10 g + 100 g)

Now, we can calculate the kinetic energy of the wooden block with the bullet after the impact using the mass of the combined system and the final velocity:

KE_combined = (1/2) * (m + M) * v_combined_final^2

Substituting the values, we get:

KE_combined = (1/2) * (10 g + 100 g) * [(10 g * 110 m/s) / (10 g + 100 g)]^2

Now, let's calculate the values.

Calculation

Using the given values, we can calculate the kinetic energy of the bullet and the kinetic energy of the wooden block with the bullet after the impact.

KE_bullet = (1/2) * 10 g * (110 m/s)^2

KE_combined = (1/2) * (10 g + 100 g) * [(10 g * 110 m/s) / (10 g + 100 g)]^2

Result

After performing the calculations, we find:

KE_bullet = 6050 J

KE_combined = 550 J

Therefore, the kinetic energy of the wooden block with the bullet after the impact is 550 J.

Conservation of Mechanical Energy

To determine whether the total mechanical energy is conserved in this case, we need to compare the initial mechanical energy with the final mechanical energy.

The initial mechanical energy is the kinetic energy of the bullet before the impact, which we calculated as 6050 J.

The final mechanical energy is the kinetic energy of the wooden block with the bullet after the impact, which we calculated as 550 J.

Since the initial mechanical energy (6050 J) is not equal to the final mechanical energy (550 J), we can conclude that the total mechanical energy is not conserved in this case.

Please let me know if I can help you with anything else.