Answer:
Explanation:
Using the law of conservation of momentum;
here;
There is a need for conservation of the total momentum that occurred before and after the collision.
So;
= mass of cart X
= mas 9f cart Y
= velocity of cart X (before collision)
= velocity of cart Y (before collision)
= velocity of cart X (after collision)
= velocity of cart Y (after collision)
So;
because the mass is identical and v represents the velocity of both carts.
Now;
= 2 m/s
= 0 ( at rest)
∴
m(2) = (2m)v
v = 1 m/s
Thus, we can see from the graphical image attached below that the velocity of X reduces to 1 m/s after collision with cart Y.
Answer:
7560 Joules
Explanation:
= Mass of first car = 
= Mass of second car = 
= Initial Velocity of first car = 0.3 m/s
= Initial Velocity of second car = -0.12 m/s
v = Velocity of combined mass
As linear momentum of the system is conserved
Energy lost is
The Energy lost in the collision is 7560 Joules
Answer:
e = 0.0898m
v = 2.07m/s
Explanation:
a) According to Hooke's law
F = ke
e is the extension
k is the spring constant
Since F = mg
mg = ke
e = mg/k
Substitute the given value
e = 1.1(9.8)/120
e = 10.78/120
e = 0.0898m
Hence it is stretched by 0.0898m from its unstrained length
2) Total Energy = PE+KE+Elastic potential
Total Energy = mgh +1/2mv²+1/2ke²
Substitute the given value
5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²
Solve for v
5.0 = 2.156+0.55v²+0.48338
5.0-2.156-0.48338= 0.55v²
2.36 =0.55v²
v² = 2.36/0.55
v² = 4.29
v ,= √4.29
v = 2.07m/s
Hence the required velocity is 9.28m/s
