1Object A is moving due east, while object B is moving due north. They collide a

Question

1Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 18.0 kg and an initial velocity of v0A = 8.10 m/s, due east. Object B, however, has a mass of mB = 28.5 kg and an initial velocity of v0B = 4.85 m/s, due north. Find the magnitude of the final velocity of the two-object system after the collision.

2Two balls with masses of of 2.3 kg and 6.2 kg travel toward each other at speeds of 14 m/s and 3.9 m/s, respectively. If the balls have a head-on inelastic collision and the 2.3-kilogram ball recoils with a speed of 7.80 m/s, how much kinetic energy is lost in the collision?

Explanation / Answer

1)Here The velocity of A = 8.10m/s
the velocity of B = 4.85m/s
Assume east = 0° and north = 90°
By conservation of momentum P, we know Pf = Pi
Since they are perpendicular, all we have to do is sum them using Pythagoras Theorem to find magntitude of Pi and Pf = Pi:
Pf = (18.0*8.10)²+(28.5*4.85)²] = 200.90 at arctan[(28.5*4.85)/(18.0*8.10)] = 43.22°
200.90kgm/s at 43.2° <<<<<<<<<<<

2)m1u1 + m2u2 = m1v1 + m2v2;

2.3*14 + 6.2*(-3.9) = 2.3*(-7.80) + 6.2*v2
32.2 - 24.18 = -17.94 + 6.2v2;

v2 = 25.96/6.2 = 4.18 m/s;
KE lost = KE(before) - KE(after);
KE lost = [(m1u1)2/2 + (m2u2)2/2] - [ (m1v1)2/2 + (m2v2)2/2]
KE lost = 810.75-496.73

K.E lost=314.01j

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