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1. A truck P, of mass m kg, is moving with speed 15 m/s towards a lorry, Q, of mass 3000 kg, which is at rest. Immediately after they collide, the speed of P is 3 m/s and its direction is reversed. The speed of Q is 9m/s after the collision. Find the mass m of P. A= the sum of the digits in your answer.

2. The forces F1=(7i-9j)N and F2=(5i+6j)N act on a particle. Find the angle, to the nearest degree that the direction of the resultant force makes with j. B= the sum of the digits in your answer.

3. A non-uniform rod PQ, of mass m and length 5H rests horizontally in equilibrium on two supports at R and T, where PR=TQ=H. The centre of mass of the rod is at K, where KT=5H/2. A particle of mass 5m/2 is placed at the midpoint of the rod. The magnitude of the normal reaction between the support at T and the rod is Fmg/12. Find F.

4. A stone is projected vertically upwards with speed 17.5m/s. The length of time for which it is at least 6.6m above its starting position is D/7 seconds. Find D.

5. A particle P of mass 4kg is projected up a fixed rough plane at an initial speed of 16m/s. The slope is inclined at an angle of 30 degrees to the horizontal. The coefficient of friction is 0.726. Find E, the distance to the nearest whole metre that the particle travels before coming to rest.

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The copied and pasted text for question 2 looks a little different, so I've attached a screen capture of how it looked to me.

When two objects collide, the quantity momentum is preserved. Momentum is velocity times mass.

So at first, the momentum is 15m. Then the momentum is -3m+9*3000.

Notice, importantly, that it's "-3" because the direction is reversed.

So you just set these equal and solve:

15m = -3m + 9*3000

18m = 9*3000

m = 1500

This means A = 6

Forces written as vectors are easy to add -- you just add them:

7i-9j + 5i+6j = 12i - 3j.

Then you want to find the angle with the vector j = 0i+1j. One way to do that is the dot product:

(12i-3j) . (0i+1j) = 0*12 + (-3)*1 = -3

The dot product obeys the rule:

|u . v| = |u| |v| cos(θ)

In this case, you get:

3 = 3√(17) cos(θ)

θ = arccos(1/√(17)) ≈ 76 °

So B = 13.

You can work out that the center of mass is H/2 away from R, whereas the other mass is right between R and T. So how much of the weight is held by the support at R? (Note: It's not just half!)

One way to think of it is to remember equal and opposite forces apply, so if you want to "flip" the system, you could consider instead two people on a see-saw. If they are balanced at different distances from the tipping point, then they must be different weights right?

This is the same -- if someone mass is balanced between these two supports, the forces work the same way. The particle contributes (5m/2)g evenly between R and T, since it is exactly between the two points, but the rod itself pushes more on R since it's closer to R. If w1 and w2 are the weights of the rod on R and T respectively, then you know:

w1 + w2 = mg

w1(H/2) = w2(5H/2)

That gives you w1=5mg/6 and w2=mg/6.

So the total weight on T is 5mg/6 + 5mg/4 = 25mg/12. So F=25.

Projectiles follow a fairly basic path, especially when thrown straight up.

The equation for such motion is derived from basic calculus, assuming the contsant acceleration "g due to gravity:

h(t) = h(0) + v(0)t + g t^2/2

On Earth g = -9.8, and in this case v(0)=17.5 is given. We should probably assume h(0)=0.

That gives us all our values:

h(t) = 17.5t - 4.9t^2

And we just solve for the two places where h(t)=6.6. The quadratic equation is easy enough, you get 3/7 and 22/7. The interval of time is 19/7, so D=19.

The coefficient of friction tells you how much the surface will slow down the moving object. The force of gravity pulls the particle downwards with force 4g (g is again the 9.8 from gravity on Earth). The component in the direction of the inclined surface is 4g cos(30°), which gives an overall force due to friction of:

0.726*4g*cos(30°) = 24.6464...

Since that is the force, the acceleration (or really, deceleration) of the object due to friction will be 6.1616. This will work counter to the velocity. Gravity also pulls against the velocity, with force g*sin(30°), which is 4.9. This means the velocity is given as a function of time to be:

v(t) = v(0) - 11.0616t

You know v(0) is 16, so t = 1.44645 when it stops.

The distance traveled is given by the integral of v(t), so:

d(t) = 16t - 11.0616t^2/2

and d(1.44645) is 11.5716.

Rounded off, E=12

A = 6

B = 13

F = 25

D = 19

E = 12

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