Momentum, Force and Conservation

[Jack Brown]

Homework Statement

Three smooth stones, A, B and C are initially at rest, in contact with each other on the smooth surface of a frozen lake (Figure 1). The masses of the stones are, to 3 significant figures, mA = 8.00 × 10^2 g, mB = 6.00 × 10^2 g and mC =2.50 × 10^2 g and the coefficient of sliding friction between the stones and the ice is μslide = 2.50 × 10^−2. An explosion between the stones causes them to fly apart across the surface of the ice. Stone A flies off due north with an initial speed of 2.40 m s^−1. Stone B flies off due east with an initial speed of 3.60 m s^−1.

(a) Calculate the speed and direction of motion of stone C immediately after the explosion. You can neglect the effects of any exhaust material from the explosion. (Remember that this is a two-dimensional problem and that you should define a suitable coordinate system for your calculation using diagrams where necessary.)

(b) If 10% of the total energy of the explosion was converted to the kinetic
energy of the stones, what was the total energy released in the explosion?

c) Draw a diagram showing all the forces acting on stone A when it is sliding across the ice after the explosion. Calculate how far it travels before it comes to rest. You can assume that the stone does not roll at any time.

Homework Equations

[/B]
a) momentum = mass x velocity
Pythagoras c^2 = a^2 + b^2
Initial momentum = final momentum

b) Kinetic = 1/2 x mass x velocity^2

c) F=ma where a = μ x g
V = u + at
s = ut + 1/2a(t^2)

The Attempt at a Solution

[/B]
a)
Momentum A = 0.8 x 2.4 = 1.92
Momentum B = 0.6 x 3.6 = 2.16
Momentum C = 0.25 x V

Using a vector triangle shown in attached picture under a)
initial momentum = final momentum
0 = momentum A + momentum B - momentum C
momentum C = momentum A+ momentum B
From vector triangle
(Mc x Vc)^2 = 1.92^2 + 2.16^2
Vc = (1.92^2)+(2.16^2) / (0.25^2) (m = 0.25)
Vc = 11.6ms-1Angle Rho = tan^-1 (1.92/2.16) = 41.6 degrees
So the correct direction is is 138.4 degrees below the x axis

b)
Kinetic energy of each disc using 1/2 m v^2
0.5 x 0.8 x 2.4^2 = 2.3
0.5 x 0.6 x 3.6^2 = 3.9
0.5 x 0.25 x 11.6^2 = 16.7

Total energy given in explosion is sum of totals x 10 = 228.6 Joules

c)
Using diagram under b)
Va = 2.4
acceleration = 9.81 x μ
V = u + at
2.4= 0 + 0.25t
t= 2.4/0.25 = 9.6 s

s = ut + 0.5 a t^2
s = 0 + 0.5 x 0.25 x 9.6
s = 11.5mPlease can you help. I struggle with this question.

 

[haruspex]

Your diagram for a) would be better with labels, A, B, C.

direction is is 138.4 degrees below the x axis

It would be clearer to say 'clockwise from the +ve x axis'. But you are given directions in terms of the compass, so I would answer in those terms.
You diagram for c) seems to show two forces and a velocity. It only asks for forces to be shown. It is a good idea to show the direction of movement, but it would be better to do that with an arrow above the diagram rather than as a line from the stone (which makes it look like a force). If it is meant to be a force it is labelled wrongly. I would also label the vertical upward force.

s = 0 + 0.5 x 0.25 x 9.6

The friction is given as

2.50 × 10^−2

 

[Jack Brown]

Apart from that, i worked it out right?

 

[haruspex]

Apart from that, i worked it out right?

Yes, looks right, though there is a slightly quicker way. You don't care about time, so use the SUVAT equation that does not involve time. (It's the same as using conservation of energy.)

 

[HalfLostAndSearching]

(a) The initial momentum of the system is zero, since all three stones are at rest. After the explosion, the momentum of stone A is 0.8 x 2.4 = 1.92 kg m/s and the momentum of stone B is 0.6 x 3.6 = 2.16 kg m/s. Since momentum is conserved, the total momentum after the explosion must also be zero. This means that the momentum of stone C is equal to the sum of the momenta of stones A and B: 1.92 + 2.16 = 4.08 kg m/s.

To find the speed and direction of stone C, we can use the Pythagorean theorem to find the magnitude of its velocity: |Vc| = √(1.92^2 + 2.16^2) = 2.77 m/s. The direction can be found using trigonometry: tan(θ) = 1.92/2.16, so θ = tan^-1(1.92/2.16) = 41.6 degrees. Since stone C is moving in the opposite direction of stone A, its direction is 180 + 41.6 = 221.6 degrees below the x-axis.

(b) The total energy released in the explosion can be found by using the kinetic energy equation: KE = 1/2 mv^2. The total kinetic energy of the system after the explosion is given by 1/2 (0.8 + 0.6 + 0.25) (2.4^2 + 3.6^2 + Vc^2) = 1.54 + 6.48 + 0.25Vc^2. Since 10% of the total energy was converted to kinetic energy, we can set this equal to 0.1 times the total energy released in the explosion: 0.1E = 1.54 + 6.48 + 0.25Vc^2. Solving for E, we get E = 22.2 J.

(c) The only force acting on stone A while it is sliding across the ice is the force of friction, which is given by F = μmg. The acceleration of stone A can be found using Newton's second law: a = F/m = μg. Since the initial velocity of stone A is