Question

A simple pendulum of length 1m has a wooden bob of mass 1kg. It is struck by a bullet of mass 10^-2 kg moving with a speed of 2oo m/s. The bullet gets embedded into the bob. Obtain the height to which the bob rises before swinging back. Take g=10 m/s^2

Answer 1

m = mass of the bullet = 0.01 kg

v = speed of the bullet before hitting the bob = 200 m/s

M = mass of the wooden bob = 1 kg

V = speed of bullet-bob combination after collision = ?

Using conservation of momentum

m v = (m + M) V

(0.01) (200) = (0.01 + 1) V

V = 1.98 m/s

h = height to which the bob of pendulum is raised

using conservation of energy

electric potential energy gained = kinetic energy lost

(m + M) g h = (0.5) (m + M) V²

V = sqrt(2gh)

1.98 = sqrt(2 x 9.8 h )

h = 0.2 m

Answer 2

Answer:

0.2 m is the height

Explanation:

Simple pendulum of length=1m

Mass=1kg

Bullet of mass=10  ^−2  kg

Speed =2×10 ^2  m/s

Height to which the bob rises before swimming back(g=10m/s  ^2 )

Now finding speed of bullet − bob combination after collison =V

using the law of conservation of momentum

mv=(m+M)V

(0.01)(200)=(0.01+1)V

V=1.98m/s

h= Height to which the bob of pendulum is raised

using conservation of energy

electric potential energy gained−

Kinetic energy lost

(m+M)gh=(0.5)(m+M)V  ^2

 

V=$\sqrt{2gh}$

 

1.98=   $\sqrt{2. 10. h}$

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h= $1.98^{2}$ ×2×10=  0.2m