A 300-kg projectile is fired from the ground with a speed of 100 m/s at a 60° an

Question

A 300-kg projectile is fired from the ground with a speed of 100 m/s at a 60° angle above the horizontal. At the highest point of its arc, it explodes into three pieces, A, B, and C, all of equal mass. Immediately after the explosion, pieces A and B are still moving with the same speed as the whole projectile had right before the explosion. However, A is moving vertically downward, and B is moving horizontally forward (i.e. the same direction as the projectile immediately prior to its explosion). Find the velocity of C after the explosion. Answer= 112 m/s, 26.6° above the horizontal

Explanation / Answer

The mass of the projectile is m = 300 kg The speed of the projection is u = 100 m/s The angle of projection is = 60o At the heighest point of the trajectory, the projectile has only horizontal component of velocity vx = ucos vx = 50 m/s The total momentum of the object before explosion is Pi = mvx Pi = 15000 kgm/s the speed of A is vA = 50 m/s vertically downward the speed of B is vB = 50 m/s horizontal Let v be the speed of C with below the horizontal The X component of the velocity of C is vcos The total momentum of the system in X direction is Pf = (100 kg)(50 m/s) + (100 kg)(vcos) Pf = 5000 + 100 vcos According to the conservation of linear momentum along horizontal direction, Pi = pf 15000 = 5000 + 100 vcos vcos = 100 According to the conservation of linear momentum along vertical direction, 0 = (100 kg)(50 m/s) + (100 kg)(vsin) vsin = -50 Therefore the magnitude of the velocity of C is v = 111.8 m/s ˜ 112 m/s The direction of the velocity of C is = tan-1(-50/100) = 26.56o ˜ 26.6o At the heighest point of the trajectory, the projectile has only horizontal component of velocity vx = ucos vx = 50 m/s The total momentum of the object before explosion is Pi = mvx Pi = 15000 kgm/s the speed of A is vA = 50 m/s vertically downward the speed of B is vB = 50 m/s horizontal Let v be the speed of C with below the horizontal The X component of the velocity of C is vcos The total momentum of the system in X direction is Pf = (100 kg)(50 m/s) + (100 kg)(vcos) Pf = 5000 + 100 vcos According to the conservation of linear momentum along horizontal direction, Pi = pf 15000 = 5000 + 100 vcos vcos = 100 According to the conservation of linear momentum along vertical direction, 0 = (100 kg)(50 m/s) + (100 kg)(vsin) vsin = -50 Therefore the magnitude of the velocity of C is v = 111.8 m/s ˜ 112 m/s The direction of the velocity of C is = tan-1(-50/100) = 26.56o ˜ 26.6o According to the conservation of linear momentum along vertical direction, 0 = (100 kg)(50 m/s) + (100 kg)(vsin) vsin = -50 Therefore the magnitude of the velocity of C is v = 111.8 m/s ˜ 112 m/s The direction of the velocity of C is = tan-1(-50/100) = 26.56o ˜ 26.6o vsin = -50 Therefore the magnitude of the velocity of C is v = 111.8 m/s ˜ 112 m/s The direction of the velocity of C is = tan-1(-50/100) = 26.56o ˜ 26.6o

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