Explosions are a fun way to learn how to apply the law of momentum to an object that starts as a single object, and after the explosion, scatters into fragments that each have their own momentum, like a firecracker. The vector sum of all the parts of the system could be added together to find the total momentum after the explosion, which equals the total momentum before the explosion. If we put a cannon on wheels, we can find the momentum change of the cannon ball and the acceleration of the cannon after it fires:
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When the cannon exploded in the problem above (in the video), the momentum of the total system is conserved. Now remember, that system is two objects: the cannon and the cannon ball. Before it exploded, the total momentum of the system is zero (nothing is moving yet). But after it fires, the total momentum of the system must still be zero for conservation of momentum. If the ball moves with 100 units of momentum to the right, then the cannon moves with 100 units of momentum to the left, so they always sum to zero.
Newton actually expressed his second law of motion in terms of momentum by stating that the rate of change of the momentum of a particle is proportional to the net force acting on the particle and is in the direction of that force.
[am4show have='p9;p58;' guest_error='Guest error message' user_error='User error message' ]
When the cannon exploded in the problem above (in the video), the momentum of the total system is conserved. Now remember, that system is two objects: the cannon and the cannon ball. Before it exploded, the total momentum of the system is zero (nothing is moving yet). But after it fires, the total momentum of the system must still be zero for conservation of momentum. If the ball moves with 100 units of momentum to the right, then the cannon moves with 100 units of momentum to the left, so they always sum to zero.
Newton actually expressed his second law of motion in terms of momentum by stating that the rate of change of the momentum of a particle is proportional to the net force acting on the particle and is in the direction of that force.
This is an example of the distributive property. We start with 0=MV+m(v+V) then, we multiply m times v and m times V. This becomes 0=MV+mv+mV. Then we collect the terms so the equation is 0 = V (M + m) + mv.
At 4:35 where did the extra little ‘m’ come from?