The man runs towards the cart. Human speed 2 m / s, trolley speed 1 m / s. The man jumps onto the cart

| May 23, 2021 | education

The man runs towards the cart. Human speed 2 m / s, trolley speed 1 m / s. The man jumps onto the cart and remains on it. What will be the speed of the cart after that if the mass of the person is 2 times the mass of the cart?

To find the value of the speed of the cart after the oncoming jump of a person n, we apply the law of conservation of momentum: Vr * (mh + mt) = Vh * mh – Vt * mt, whence we express: Vcht = Vh * mh – Vt * mt / (mh + mt).

Variables: mh (human weight) = 2mt (trolley weight); Vh – running speed of a person (Vh = 2 m / s); Vt is the initial speed of the bogie (Vt = 1 m / s).

Calculation: Vch = Vh * mh – Vt * mt / (mh + mt) = 2 * 2mt – 1 * mt / (2mt + mt) = 3mt / 3mt = 1 m / s.

Answer: After a person jumped, the cart will start moving in the opposite direction at a speed of 1 m / s.



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