In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.0 cm. A second projectile causes the the pendulum to swing higher, to h2 = 5.5 cm. The second projectile was how many times faster than the first?

Verified Answer :

Assuming you mean " how fast at the bootom of the swing", since the speed of a pendulum is constantly changing. The key with pendulums is the transfer of energy from potential to kinetic. You can use the next formulas 
m1∗g∗(2.0cm)=12m1v2
The speed of the bottom is equal to
v=(√2∗g∗2.0cm)
If a second pendulum reaches twice as high, substitute twice the height into the same equation

The second projectile was about 1.7 times faster than the first.

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Further explanation

Let's recall Kinetic Energy Formula as follows:

\(Ek = \frac{1}{2}mv^2\)

Ek = Kinetic Energy ( Joule )

m = mass of the object ( kg )

v = speed of the object ( m/s )

Let us now tackle the problem !

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Given:

height of pendulum 1 = h₁ = 2.0 cm

height of pendulum 2 = h₂ = 5.5 cm

Asked:

speed of projectile 2 = v₂ = ?

Solution:

Firstly , we will use Conservation of Energy as follows:

\(Ek = Ep\)

\(\frac{1}{2}(m+M)v^2 = (m+M)gh\)

\(\frac{1}{2}v^2 = gh\)

\(v^2 = 2gh\)

\(v = \sqrt{2gh}\)

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Next, we will use Conservation of Momentum to find the speed of projectile:

\(mv' = ( m + M )v\)

\(v' = (1 + \frac{M}{m})v\)

\(v' = (1 + \frac{M}{m}) \sqrt{2gh}\)

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Finally , we will compare speed of projectiles of the two experiments as follows:

\(v_1' : v_2' = (1 + \frac{M}{m})\sqrt{2gh_1} : (1 + \frac{M}{m})\sqrt{2gh_2}\)

\(v_1' : v_2' = \sqrt{h_1} : \sqrt{h_2}\)

\(v_1' : v_2' = \sqrt{2.0} : \sqrt{5.5}\)

\(v_2' = \sqrt{\frac{5.5}{2.0}} \times v_1'\)

\(v_2' = 1.7 v_1'\)

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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