Question

Cнаряд летевший горизонтально со скоростью 80м-с,разорвался на высоте 30м на 2 разных осколка.1 осколок упал точно под местом взрыва. какова будет скорость 2 оскалка и в каком направлении он станет двигаться?

Answer 1

mV=m₁V₁+m₂V₂

m₁ упал наземлю точно под местом взрыва, т.е. перпендикулярно движению снаряда, тогда импульс m₂ будет направлен в противоположную сторону, т.е. вверх, тудаже будет направленна V₂

V₂=(mV-m₁V₁)/m₂

Answer 2

Calculation of the Second Fragment's Speed and Direction

To calculate the speed and direction of the second fragment, we need to consider the conservation of momentum. The momentum of an object is defined as the product of its mass and velocity. In this case, we have two fragments resulting from the explosion of a projectile.

Let's denote the first fragment as fragment 1 and the second fragment as fragment 2. We know that fragment 1 fell exactly below the explosion site, which means its velocity is zero in the horizontal direction. Therefore, the momentum of fragment 1 in the horizontal direction is zero.

The momentum of an object is conserved unless there are external forces acting on it. In this case, we can assume that there are no external forces acting on the fragments after the explosion. Therefore, the total momentum before the explosion should be equal to the total momentum after the explosion.

Since the first fragment has zero momentum in the horizontal direction, the total momentum after the explosion is equal to the momentum of the second fragment. Let's denote the mass of the second fragment as m2 and its velocity as v2.

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

m1 * v1 + m2 * v2 = 0

where m1 is the mass of the first fragment and v1 is its velocity. We know that m1 = 0 (since the first fragment has zero mass) and v1 = 0 (since the first fragment has zero velocity in the horizontal direction). Therefore, the equation simplifies to:

m2 * v2 = 0

Since m2 is not zero, we can conclude that v2 must be zero. This means that the second fragment will have zero velocity in the horizontal direction. In other words, it will not move horizontally.

To summarize: - The speed of the second fragment is zero. - The second fragment will not move horizontally.

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