Question

A bullet of mass 45 g hits a block of mass 3.66 kg with a speed v. After the collision, the bullet emerges at the other end of the block with a speed of 87 m/s, and the pendulum swings up over an angle of 38°. If the string of the pendulum is 3.6 m, what was the initial speed v of the bullet? st

Answer 1

Solution : - mas = m = of bullet 45x10 45g 3M = 3.60 Ry mas of bloch initial velocity of bulket = v final velocity of that - V₂ = 87 m/s Q = 389 b= = 3.6m. length of steeing N e-h e B from DOAB ech coso e Ih dah= l coso h= l-l coso ha e ci caso a m M - ell-coso) ☆ D. 763 m ch= 3.6 [ Cosse] ha 0.763 m Gain in Potential Energy Loss in Kinetic bullet by energy by beloch - m [v- Vin] Mg h -( v. - V.2) = 2 Mych = 2*3.6649-8X 0.763 -(v. ?- 12) = 1.22 x 10 us x 10 3 3 2 Vz V, 23 + 1.22x103 87+1.22X10 3

2. 1,2 = 7.6 X10% + 1.22 x103 2 U? = 8.82 X 103 V 93.92 m/s to initial velocity of bullet deullet was 93.92 mle