Final answer:
To calculate the magnitude and direction of the velocity of the 0.400-kg object after the collision, we will use the principle of conservation of momentum. The magnitude of the velocity of the 0.400-kg object after the collision is 1.25 m/s, and it makes an angle of 45º with the x-axis.
Step-by-step explanation:
To calculate the magnitude and direction of the velocity of the 0.400-kg object after the collision, we will use the principle of conservation of momentum. In this case, we can assume that there are no external forces acting on the system. Since the 0.250-kg object is initially moving at 2 m/s and the 0.400-kg object is stationary, the initial momentum of the system is given by:
P_initial = m1 * v1 + m2 * v2 = (0.250 kg)(2 m/s) + (0.400 kg)(0 m/s) = 0.5 kg.m/s
After the collision, the 0.250-kg object emerges at 1.5 m/s at an angle of 45.0º with its incoming direction. We can resolve the velocity of the 0.250-kg object into its components using trigonometry:
v1x = v1 * cos(angle) = (1.5 m/s) * cos(45º) = 1.06 m/s
v1y = v1 * sin(angle) = (1.5 m/s) * sin(45º) = 1.06 m/s
Since the 0.400-kg object is initially stationary, its initial momentum is zero:
P_initial = m1 * v1 + m2 * v2 = 0 kg.m/s + (0.400 kg)(v2) = 0.5 kg.m/s
Solving for v2:
v2 = (0.5 kg.m/s) / (0.400 kg) = 1.25 m/s
So the magnitude of the velocity of the 0.400-kg object after the collision is 1.25 m/s. To determine the direction, we can calculate the angle it makes with the x-axis:
tan(theta) = v1y / v1x = (1.06 m/s) / (1.06 m/s) = 1
theta = arctan(1) = 45º