We know that a=vf_vi/t equals equation "a" . Where a is the acceleration of the body , vf is the final velocity , vi is the initial velocity and t is equal to time . Since vi equals o m/s , vf equals to 60 m/s and t equals 10 s. Put in equation "a". a=60-0/10 =6m/s2
Electric force depends on the charge and the strength of the electric field. The equation that relates the three:
F = Eq where q is the charge and E is the electric field strength.
Answer: It's hard to say without characterizing the collision. But it will be either A if the collision is totally in-elastic, or B if the collision is totally elastic. It could be anywhere in between for partially elastic collisions.
Explanation:
momentum is conserved, so initial system momentum will be left to right.
The velocity of the center of mass is 50(5) / 550 = 0.4545... m/s
In an elastic collision, the lead ball will move off at twice that speed or 0.91 m/s to the right.
The steel ball will bounce back and move away at 0.91 - 5 = -4.1 m/s . The negative sign indicates the steel ball has reversed course and has negative momentum
In a totally in-elastic collision, both balls would move to the right at 0.45 m/s. The steel ball will still have positive momentum.
We have: K.E. = 1/2 mv²
Here: m = 50 g = 0.05 Kg
v = 4 m/s
Substitute their values,
K.E. = 1/2 * 0.05 * 4²
K.E. = 1/2 * 0.05 * 16
K.E. = 0.4 J
In short, Your Answer would be 0.4 Joules
Hope this helps!
