Answer:
2.06 m/s
Explanation:
From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.
Momentum before collision
Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5
Momentum after collision
The momentum after collision will be given by (9+27)*0.9=32.4
Relating the two then 9v+13.5=32.4
9v=18.5
V=2.055555555555555555555555555555555555555 m/s
Rounded off, v is approximately 2.06 m/s
I would say its a positive cgarge
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton,
= 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E

where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd




Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.