Question

Конькобежец массой 63 кг стоя на коньках на льду бросает камень массой 2 кг со скоростью 3 м.с. под углом 30 гр.к горизонту.определите скорость конькобежца после броска.

Answer 1

найдем проекцию скорости шайбы на горизонтальную ось 

Vx= v*cos (30)= 1,5* (корень из 3)

по закону сохранения импульса 

Р1=Р2

т.к. в начальное положение скорость была равно 0 то Р1=0

 в конечный моменты  скорости направлены в противоположные стороны  

тогда 

0=Mv1-mv2  

Mv1=mv2

v2= Mv1/m= (2*1,5*(корень из 3))/63= (корень из 3 ) /21

Answer 2

Calculation of the Speed of the Speed Skater after Throwing the Stone

To determine the speed of the speed skater after throwing the stone, we can use the principle of conservation of momentum. According to this principle, the total momentum before the throw is equal to the total momentum after the throw.

The momentum of an object is calculated by multiplying its mass by its velocity. In this case, we have the following information:

- Mass of the speed skater (m1): 63 kg - Mass of the stone (m2): 2 kg - Initial velocity of the stone (v2): 3 m/s - Angle of the throw (θ): 30 degrees

To calculate the speed of the speed skater after the throw, we need to find the initial velocity of the speed skater (v1) before the throw.

Using the conservation of momentum equation:

m1 * v1 + m2 * v2 = (m1 + m2) * vf

where vf is the final velocity of the speed skater and the stone after the throw.

Since the speed skater is initially at rest (v1 = 0), the equation simplifies to:

m2 * v2 = (m1 + m2) * vf

Substituting the given values:

2 kg * 3 m/s = (63 kg + 2 kg) * vf

Simplifying the equation:

6 kg m/s = 65 kg * vf

Dividing both sides of the equation by 65 kg:

vf = 6 kg m/s / 65 kg

Calculating the value:

vf ≈ 0.0923 m/s

Therefore, the speed of the speed skater after throwing the stone is approximately 0.0923 m/s.