1. A neon atom ( m =20.0u) makes a perfectly elastic collision with another atom

Question

1. A neon atom (m=20.0u) makes a perfectly elastic collision with another atom at rest. After the impact, the neon atom travels away at a 32.2 angle from its original direction and the unknown atom travels away at a -53.0 angle.

What is the mass (in u) of the unknown atom? [Hint: You can use the law of sines.]

2. A football is kicked at ground level with a speed of 18.9 m/s at an angle of 31.4 to the horizontal.

How much later does it hit the ground?

Express your answer using three significant figures and include the appropriate units.

3. A falling stone takes t = 0.34 s to travel past a window 2.2 m tall

From what height above the top of the window did the stone fall?

Express your answer to two significant figures and include the appropriate units.

4. You drop a 16-g ball from a height of 1.5 m and it only bounces back to a height of 0.89 m .

What was the total impulse on the ball when it hit the floor? (Ignore air resistance.)

Express your answer to two significant figures and include the appropriate units. Enter positive value if the impulse is upward and negative value if the impulse is downward.

Explanation / Answer

= 31.4
Vi= 18.9 m/s

get Vy

Vx = cos * Vi
Vx = cos 31.4 * 18.9m/s
= 16.13 m/s

tflight =2(Vy)/g
= 2(9.847 m/s) /9.8 m/s^2
= 2.007 s

Range = Vx*tflight
= 16.13 m/s * 2.007s
= 32.38  m

Time to hit the ground: 2.007 s
Range: 32.38 m

3) Initial velocity a( t t= 0 ) = Vo = 0
Velocity at top of the Window = V1 and velocity at the bottom of window = V2
S12 = 2.2 = V1 t + 1/2 g t^2 => 2.2 = V1 (0.34) + 4.9 (0.34)^2 => V1= 4.8 m/s
V1 = 4.8 = Vo + g t = 0 + 9.8 t => t = 0.49 s This is the time from start to top of window.
S (from launch point to top of window) = 0 + 1/2 (9.8) (0.49)^2 = 1.l8 m
Total distance = 2.2 + 1.18 = 3.38 m

4) impulse =change in momentum

this means we calculate the momentum of the ball just before impact and just after; we know the mass, so we find the speeds

the ball falls for 1.5m and will achieve a speed given by energy conservation:

1/2 mv^2 = mgh => v=sqrt[2gh]=5.42m/s

since it rises only to 0.89 m, we calculate the initial speed after impact from the same equation and get
v(after)=sqrt[2*9.81m/s/s*0.89m]=4.179m...

now, remember that momentum is a vector, so that the momentum down has one sign (let's call it negative), and the momentum up has a positive sign, so we have

impulse = delta (mv) = m delta v=0.016kx(4.179m/s - (-5.42m/s) = 0.154 kgm/s

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