Question

масса скейта 5 кг, а масса школьника 25 кг. Скейт стоит, а школьник прыгает на него со скоростью 8 м/с. С какой скоростью поедет школьник на скейте, если удержится от падения?

Answer 1

Ответ:

Пусть ось х направлена вдоль изначального движения мальчика и скейтборда.

Запишем закон Сохранения импульса в проекции на ось х

M-мальчик

m-скейт

V1-скорость мальчика (после того как выпрыгнул)

V2-скорость скейта (после того как мальчик спрыгнул)

(M+m)*V0=-M*V1+mV2

V2= \frac{(M+m)V0+M*V1}{m}= \frac{(30+10)*0.5+30*1}{10} =5V2=

m

(M+m)V0+M∗V1

=

10

(30+10)∗0.5+30∗1

=5 м/с

Answer 2

Calculation of the Schoolboy's Speed on the Skateboard

To calculate the speed at which the schoolboy will travel on the skateboard, we can use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, provided no external forces act on the system.

In this case, we can consider the schoolboy and the skateboard as a system. Before the schoolboy jumps onto the skateboard, the total momentum is zero since both the schoolboy and the skateboard are at rest. After the schoolboy jumps onto the skateboard, the total momentum will still be zero since no external forces are acting on the system.

The equation for momentum is given by:

Momentum = mass × velocity

Let's denote the mass of the skateboard as m1 (5 kg) and the mass of the schoolboy as m2 (25 kg). The initial velocity of the schoolboy before jumping onto the skateboard is v2 (8 m/s). The final velocity of the schoolboy and the skateboard after the jump is v (which we need to find).

According to the conservation of momentum, we can write the equation:

m1 × 0 + m2 × v2 = (m1 + m2) × v

Simplifying the equation, we get:

m2 × v2 = (m1 + m2) × v

Now we can substitute the given values into the equation and solve for v:

25 kg × 8 m/s = (5 kg + 25 kg) × v

200 kg·m/s = 30 kg × v

Dividing both sides of the equation by 30 kg, we get:

v = 200 kg·m/s / 30 kg

Simplifying the equation, we find:

v ≈ 6.67 m/s

Therefore, if the schoolboy manages to maintain his balance and not fall off the skateboard, he will travel at a speed of approximately 6.67 m/s.