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A $\\var{m1}$ kg mass travelling at $\\var{u1}$ ms-1 collides into a mass $\\var{m2}$ kg moving in the same direction $\\var{u2}$ ms-1.

\n

Give your answers to the appropriate number of significant figures.

", "advice": "

B) Use conservation of linear momentum to determine the final velocity.

\n

C) $KE=\\frac{1}{2}mv^2$

\n

See section 3.24 (pg 92) in Advanced Physics by Adams and Allday.

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Which statement defines an inelastic collision?

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Kinetic energy is not conserved, but momentum is conserved.

", "

Gravitational potential energy is converted into kinetic energy.

", "

Both momentum and kinetic energy are conserved.

", "

Momentum is not conserved, but kinetic energy is conserved.

", "

Both momentum and kinetic energy are not conserved.

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What is the velocity of the two masses immediately after the collision if they lock together?

\n

[[0]] ms-1  

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How much kinetic energy is transferred to other forms in the collision?

\n

[[0]] kJ    

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You throw a $\\var{m}$ kg ball straight up in the air, giving it an initial velocity of $\\var{u}$ ms-1.

\n

Take $g$ = 9.81 ms-2 and give your answers to three signficant figures.

", "advice": "

Part a) At the point of maximum height the ball will have transferred all its' kinetic energy to gravitational potential energy, hence:

\n

$\\frac{1}{2}mv^2=mgh$

\n

Part b) i) Sum the change in graviational potential and kinetic energies, then divide by the change in height.

\n

           ii) Sum the initial and final gravitational potential and kinetic energies, rearrange for the unknown $v$ and solve.

\n

See section 3.13 (pg 70) in Advanced Physics by Adams and Allday.

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Use conservation of energy to calculate how high it goes.

\n

[[0]]m

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You have not given your answer to the correct precision.

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Suppose that your hand moves up $\\var{hand_h}$ m while giving the ball its upward velocity of $\\var{u}$ ms-1.

\n

Assuming that your hand exerts a constant upward force on the ball, what is the magnitude of that force?

\n

[[0]]N

\n

What is the speed of the ball at a point $\\var{point_h}$ m above the point where it leaves your hand?

\n

[[1]]ms-1

\n

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Two carts A and B, with a compressed spring between them, are pushed together and held at rest, as shown in the figure. The spring is not attached to either cart. The carts are then released.

\n

Give your answers to two decimal places.

\n

\n

\n

", "advice": "

See section 3.22 (pg 88) in Advanced Physics by Adams and Allday.

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The graph below shows how the force, F, exerted by the spring on the carts varies with time, t, after release.

\n

\n

When the spring returns to its unstretched length, and the unattached spring drops down, cart A is moving at 0.60 ms-1.

\n

Calculate the impulse given to each cart by the spring as it expands:[[0]] Ns.

\n

Calculate the mass of cart A:[[1]] kg.

\n

State the final momentum of the system at the instant the spring drops away:[[2]] kgms-1

\n

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Is a little restrictive on the formatting - requires minuses in set places that the question allows to cancel… recommend use as practice not assessed

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A snooker ball moving with speed  vA = {v} ms-1 in the +x direction strikes an equal mass ball B which is initially at rest. The two balls are observed to move off at 45o above and below the x axis.

\n

We want to find the speeds of the two balls after the collision.

\n

\n

 

\n

Fig. A collision in two dimensions between snooker balls A and B, of equal mass.

\n

Note that we use the prime symbol (') here and in the answers to denote v and $\\theta$ after the collision.

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Speed of the snooker ball A before the collision.

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Firstly, complete the follow equations for momentum in the x and y directions. 

\n

\n

For the x direction: mvA = m[[0]]A  [[1]](45o) + m[[0]]B  [[1]](-45o)

\n

\n

and for the y direction: 0 = [[3]]A  [[2]](45o) + [[3]]B  [[2]](-45o)

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Bearing in mind that the two balls have equal masses and also noting that sin(-$\\theta$) = -sin($\\theta$), cos(-$\\theta$) = cos($\\theta$), complete the following expressions:

\n

v'B = [[2]]A  [[0]](45o)/[[1]](45o) = v'A

\n

vA = v'A[[3]](45o) + [[4]]B  [[3]](45o) = 2[[4]]A  cos(45o)

\n

\n

\n

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Calculate v'A = v'B = [[0]] ms-1 

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In an experiment using a linear air track a $\\var{m1}$ kg rider travelling to the right at $\\var{u1}$ ms-1 collides with a second, stationary rider of mass $\\var{m2}$ kg. Find the velocities v1 and v2 of the two riders if: a) the collision is perfectly elastic; and b) the velocity v if the collision is totally inelastic.

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From the question, we know:  $m_1=\\var{m1}$ kg,  $m_2=\\var{m2}$ kg,  $u_1=\\var{u1}$ ms-1,  and  $u_2=\\var{u2}$ ms-1.

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Give your answers to two significant figures.

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See section 3.24 (pg 92) in Advanced Physics by Adams and Allday.

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Hint: If the collision is perfectly elastic, both momentum and kinetic energy are conserved.

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first rider: $v_1 =$ [[0]] ms-1

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second stationary rider: $v_2 =$ [[1]] ms-1

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Hint: If the collision is totally inelastic, only linear momentum is conserved.

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$v =$ [[0]] ms-1

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