Momentum(assignment question)plz help?

Question

Momentum(assignment question)plz help?
Showing all steps, and solutions, thanks
realizing that he could not drive up a 30*, ice covered hill becasue there was no friction, sir issic Newton had stopped his cart, of total mass 500 kg, at the bottom. He was struck in the rear by a London stage coach, of total mass 1500 kg, travelling at 20 m/s. The two vehicles stuck together, with nothing breaking loose, and slid up the hill in a straight line. How far up the slpoe did the wreckage get before coming to rest?
doing his question would really help me out, thanks.

Answer 1

first we use conservation of momentum to find the speed of the combined mass immediately after impact, then we use that speed to use energy conservation to find the height of the combined cart up the hill....in effect, this is a ballistic pendulum experiment



the momentum before collision = 1500kg x 20 m/s = 30,000kgm/s



after collision, the momentum = 2000kg x V where V is the speed after collision



30,000kgm/s = 2000V => V= 15m/s



now we can use energy conservation to find the height of the combined mass above the ground



the KE at the bottom of the hill equals the PE at the highest point on the hill, so



1/ m v^2 = m g h



or h = v^2/ 2 g = 15^2/2*9.8 = 11.5m



this is the height above the ground, the distance up the slope is given by



slope distance = height/sin 30 = 23m

Answer 2

Mark -

Sir Issic was stopped. He brought no energy to the party. The stagecoach had momentum of 1500 kg x 20 m/s. After the collision the system's momentum was the same as before... only slower:

20m/s x 1500kg = ? m/s x 2000kg (Right? The two stayed together, the masses combined....)

There was a 200kg mass moving at 15 m/s.

Let's translate that into kinetic energy: KE = 1/2 mv^2; KE = 1/2*2000*15^2; KE = 225,000 joules.

The question becomes: How high can 225,000 joules lift a mass of 2000 kg?

We must find weight of 2000 kg. Weight = mass x gravity, W = 2000 x 9.8, W = 19,600 N. Now that's the force which must be exerted to lift the cart + stage coach. ( We can ignore the slope because, as there was no friction, the only effective work which was done was straight up... because the only other force in this problem (the weight) was straight down.)

Work = Force x Distance.

225,000 joules = 19,600 Newtons x Distance, Distance = 11.48 m ≈ 11 1/2 meters

Answer 3

Momentum of the system at the bottom = 1500x20+500x0 = 30000

Momentum of the system when it stops after covering some distance = 0

Change in momentum = 30000



Weight of the two bodies = 1500x10 + 500x10 = 15000+5000 = 20000 N (taking g=10)

Weight acting along the slope = 20000xSin 30 = 20000x0.5=10000 N

So, the force acting (downwards) on the system during the motion up the slope = 10000 N

F = ma i.e. a = F/m = 10000/2000 = -5 m/sec^2

Now, we know that the rate of change of momentum = Force

Change of momentum/Time = Force

30000/t = 10000

t = 30000/10000 = 3 sec



Now, S = Vi x t + 1/2 x a x t^2

= 20 x 3 + 0.5x(-5)x3x3 = 60 - 22.5 = 37.5 meter



So, it will move 37.5 m up the slope before coming to rest<<<<<<<<<<<<<<<<<<<<