Force acting during collision is internal so momentum is conserve
so (initial momentum = final momentum) in both directions
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.
Let Vx is and Vy are final velocities of car in +x and +y direction respectively.
initial momentum in +ve x (east) direction = final momentum in +ve x direction (east)
- 750*25 + 1150*0 = (750+1150)
Vx
initial momentum in +ve y (north) direction = final momentum in +ve y direction (north)
750*0 - 1150*5 = (750+1150)
Vy
from here you can calculate Vx and Vy
so final velocity V is
<span>V=<span>(√</span><span>V2x</span>+<span>V2y</span>)
</span>
and angle make from +ve x axis is
<span>θ=<span>tan<span>−1</span></span>(<span><span>Vy</span><span>Vx</span></span>)
</span><span>
kinetic energy loss in the collision = final KE - initial KE</span>
Answer:
Da=(1/4)Db
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
When s = Da, t = t

When s = Db, t = 2t

Dividing the two equations

Hence, Da=(1/4)Db
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Explanation:
Given that,
Charge on a spherical drop of water is 43 pC
The potential at its surface is 540 V
(a) The electric potential on the surface is given by :

r is the radius of the drop

(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,

Now the charge on the new drop is 2q. New potential is given by :

Hence, the radius of the drop is
and the potential at the surface of the new drop is 856.79 V.
Answer:(a)0,(b)1.061 kg-m/s
Explanation:
Given
mass of ball is 0.10 kg
Initial speed is 15 m/s
Maximum height reached by ball is h

final velocity =0


thus momentum of ball at maximum height is 0 as velocity is zero
For halfway to maximum height

where 


v=10.61 m/s
Thus its momentum is
