F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
(a) The momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.
(b) The size of momentum is 93500 kgm/s and it will be directed towards South.
Explanation:
The mass of the truck moving due south is given as 4000 kg and the speed is 20 m/s. Similarly, the mass of the car moving due north is 1350 kg and the speed is 10 m/s.
(a) Then the momentum of each vehicle can be obtained by the product of mass with their respective speed.
Similarly, the momentum of car will be
So, the momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.
(b) Since, after collision, the vehicles stick together, the momentum after collision will be equal to the total momentum of both the vehicles before collision. This is because, it will obey conservation of momentum.
Momentum of vehicles after collision = total momentum before collision
Momentum after collision = 80000+13500 = 93500 kgm/s.
The direction of the vehicles after collision will be towards south as the mass and speed of the truck is greater than car. So the impact or force exerted on the car by the truck will be greater and thus both the vehicles will be directed towards south after collision.
Thus, the size of momentum is 93500 kgm/s and it will be directed towards South.