Suppose the following experiment is performed. A 0.250-kg object (m1)(m1) is sli

Question

Suppose the following experiment is performed. A 0.250-kg object (m1)(m1) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg (m2)(m2). The 0.250-kg object emerges from the room at an angle of 45o45o with its incoming direction.The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (v2(v2 and 2)2) of the 0.400-kg object after the collision.

Explanation / Answer

using law of conservation of momentum

along X-axis

m1*u1 = m1*v1*cos(theta) + m2*v2*cos(alpha)

0.25*2 = 0.25*1.5*cos(45) + (0.4*v2*cos(alpha))

v2*cos(alpha) = 0.587 m/sec

again along Y-axis

0 = 0.25*1.5*sin(45) +(0.4*v2*sin(alpha))

v2*sin(alpha) = 0.662

tan(alpha) = 0.662/0.587 = 1.127

alpha = tan^(-1)(1.127) = 48.4 degrees with the original direction

v2*cos(48.4) = 0.587

v2 = 0.587/cos(48.4) = 0.884 m/sec

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