A simple pendulum of length 1 m has a wooden bob of mass 1 kg . It is struck by a bullet of mass 10-2 kg moving with a speed of 2 × 102 ms-1 . The height to which the bob rises before swinging back is (Take g = 10 ms-2 )

Question

A simple pendulum of length 1 m has a wooden bob of mass 1 kg . It is struck by a bullet of mass 10-2 kg moving with a speed of 2 × 102 ms-1 . The height to which the bob rises before swinging back is (Take g = 10 ms-2 ) (A)  0.2 m  (B)  0.6 m  (C)  8 m  (D)  1 m

Tardigrade Admin · Accepted Answer

Momentum of bullet = 10-2 × 2 × 102 = 2 kg ms-1 . Let the combined velocity of the bob + bullet = v . Momentum of bob + bullet = (10-2 + 1)v = 1.01 v . By conservation of momentum, 1.01 v = 2 kg ms-1 or v = (2/1.01)=1.98 ms-1 . By conservation of energy (1/2)(M + m)v2 = (M + m)gh or = h = (v2/2 g) = ((1.98)2/2 × 10) = 0.19 ≃ 0.2 m

Q. A simple pendulum of length 1 m has a wooden bob of mass $1 k g$ . It is struck by a bullet of mass $1 0^{- 2} k g$ moving with a speed of $2 \times 1 0^{2} m s^{- 1}$ . The height to which the bob rises before swinging back is
(Take $g = 10 m s^{- 2}$ )

$0.2 m$

$0.6 m$

$8 m$

$1 m$

Q. A simple pendulum of length 1 m has a wooden bob of mass 1kg . It is struck by a bullet of mass 10−2kg moving with a speed of 2×102ms−1 . The height to which the bob rises before swinging back is (Take g=10ms−2 )

0.2m

0.6m

8m

1m

Q. A simple pendulum of length 1 m has a wooden bob of mass $1 k g$ . It is struck by a bullet of mass $1 0^{- 2} k g$ moving with a speed of $2 \times 1 0^{2} m s^{- 1}$ . The height to which the bob rises before swinging back is
(Take $g = 10 m s^{- 2}$ )

$0.2 m$

$0.6 m$

$8 m$

$1 m$