Question

Два тела массами 3 и 5 кг, которые двигались навстречу друг другу со скоростями 4 м/с, подверглись абсолютно неупругому столкновению. Определите (в джоулях) сколько механической энергии перешло во внутреннюю.

Answer 1

M2*V-m1*V=(m1+m2)*U
U=V*(m2-m1)/(m1+m2)=4*(5-3)/(5+3)=1 м/c
Q=(m1*V^2/2+m2*V^2/2)-(m1+m2)*U^2/2=0,5*(m1+m2)*(V^2-U^2)=0,5*(3+5)*(16-1)=60 Дж

Answer 2

Calculation of Mechanical Energy Transfer in an Inelastic Collision

To determine the amount of mechanical energy transferred to internal energy during an inelastic collision between two bodies, we need to consider the conservation of momentum and the conservation of kinetic energy.

Given: - Mass of the first body (m1) = 3 kg - Mass of the second body (m2) = 5 kg - Initial velocity of the first body (v1) = 4 m/s - Initial velocity of the second body (v2) = -4 m/s (since they are moving towards each other)

To find the final velocities of the bodies after the collision, we can use