Answer:
1)4.7334J
2)225.4m/s
Explanation:
v= the Velocity of both the bullet and the block after collision=?
H= Height of the bullet along circular arc= 10cm=0.1m
g= acceleration due to gravity= 9.81m/s^2
R= Radius of the circular arc= 18cm= 0.18m
m= Mass of the bullet= 30g= 0.03kg
M= Mass of the block = 4.8 kg
Using the law of conservation of energy
Potential energy of the system= Kinectic energy of the system
1/2 mv^2= mgh..............eqn(1)
But we have two mass m and M
We can write eqn(1) as
0.5(m+M)v^2= (m+M)gh ...........eqn(2)
If we make "v" subject of the formula we have
v = √2gh
Then substitute the values we have
= √2 x 9.81 x 0.1 = 1.40m/s
1) We can now calculate the total energy of the system after collision as
KE = 1/2(m+M)v^2
= 1/2 x (0.03+4.8) x (1.40)^2
KE = 4.7334J
Hence, the total energy of the composite system at any time after the collision is 4.7334J
2)to determine the initial velocity of the bullet.
From law of momentum conservation, which can be expressed as
m1u1+m2u2=(m1+m2)v
Where the initial Velocity of the bullet u1= ?
Final velocity of the bullet = 0
the Velocity of both the bullet and the block after collision=v= 1.40m/s
(0.03×u1) +(u×0)= (4.8+0.03)1.4
0.03u1=6.762
U1=225.4m/s
Hence, the initial velocity of the bullet is 225.4m/s
] water with ~18 MeV protons which are high energy in nature. The bombarding is achieved using a cyclotron or an accelerator.