Question

PROBLEM 17,101 A 45-g bullet is fired with a velocity of 400 m/s at 0 = 30° into a 9-kg square panel of side b = 200 mm. Knowing that h=150 mm and that the panel is initially at rest, determine (a) the velocity of the center of the panel immediately after the bullet becomes embedded, (b) the impulsive reaction at A, assuming that the bullet becomes embedded in 2 ms.

please do not just copy solution manual..

Answer 1

i.e. impulsive reaction at A is in y-direction of magnitude 4500N.

Solution : - Ay de - mo- Axe 45g=0.045 kg s 24 b - rm = vo guy > Vo = 4comis 301 ज b h •m b 7o. 300 Къ - b = 200mm I ofter strike Before strike - h= 150mm At = 2x10-38 (A) ) For velocity of Centroid abbly momenhem conservation ine-direction 2 momentum after strike = momentum before strike = mxot mo Vocos30° (m + mo) v (9 +0.045)v = 0.045x 400 x COS 30° 5) v = 1.723 mis after the bullet has embeded > velocity of Centered immediately 1.723 mis = (b) For reaction at À find À find both horizontal & vertical component of Impulsive reaction. (Ax, Ay) change Abex At (m+mo) V – mo Vo Ces 300 =) | Ax Ax = 0 In x-direction > Impulsive reaction in x-direction in momentum in e-direction. → Im y-direction :- Ayx At = (mimo) xo - mot-vosin 30 ) 5) Ayx 2x10-3 = +0.045 x 400 x sin 30° » [Ay 45 con Impulsive reaction at À =