Car a with mass 1,783 kg collides with stationary 1600 kg car b. they become locked together after the collision and move with s
1 answer:
The initial speed of car A is 15.18 m/s.
Momentum is defined as mass in motion. If there are two objects (the two objects in motion or only one object in motion and the other in stationary) that collide and no other forces work in the system, the law of momentum conservation applies in the system.
p=p'
pa+pb = pa'+pb'
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
- ma = mass of object A (kg) = 1,783 kg
- mb = mass of object B (kg) = 1,600 kg
- va = speed of object A before collides (m/s)
- va' = speed of object A after collides (m/s) = 8 m/s
- vb = speed of object B before collides (m/s) = 0 m/s
- vb' = speed of object B after collides (m/s) = 8 m/s
- p = momentum before collision (Ns)
- p' = momentum after collision (Ns)
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
(1,783×va) + (1,600×0) = (1,783×8) + (1,600×8)
(1,783×va) + 0 = 14,264+12,800
(1,783×va) = 27,064
va = 15.18 m/s
Learn more about The law of momentum conservation here: brainly.com/question/7538238
#SPJ4
You might be interested in
Answer:
Explanation:
The volume of a sphere is:
V = 4/3 * π * a^3
The volume charge density would then be:
p = Q/V
p = 3*Q/(4 * π * a^3)
If the charge density depends on the radius:
p = f(r) = k * r
I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.
Since p = k*r
Q = p*π^2*r^3 / 2
Then:
p(r) = 2*Q / (π^2*r^3)