A perfectly (non-head on) elastic collision occurs between a 0.25kg object moving east at 4m/s and a stationary 0.3kg object. Following the collision the 0.25kg object is moving at 2m/s.
a) What is the speed of the 0.3kg object after the collision?
I did:
What formula do you use?
I used m1v1^2 = m1v1'^2 + m2v2'^2
and got 3.16m/s but I'm not sure if that's right.
b) You must select suitable directions of motion for the 2 objects after this collision, which is not head-on. Calculate the direction of motion for one of the 2 objects after the collision.
The conservation of energy applies, you did it correctly in a).
For b), you have to use the result in a), and then set up a coordinate system E,N. The N components of both velocities have to add to zero, and the E components have to add to 4m/s. Use trig to get the components of each, assume some angle N of E, and S of E, theta1 and theta2. The algebra is a bit messy, but it works out.
a) To find the speed of the 0.3kg object after the collision, you can use the law of conservation of momentum. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.
The formula for momentum (p) is given by:
p = mv
Given the mass (m1) and initial velocity (v1) of the 0.25kg object, and the final velocity (v1') of the same object, we can calculate the initial momentum of the 0.25kg object as:
p1_initial = m1 * v1
Similarly, the final momentum of the 0.25kg object is:
p1_final = m1 * v1'
Then, using the conservation of momentum, we have:
p1_initial + p2_initial = p1_final + p2_final
Since the second object is stationary (v2 = 0), the initial momentum of the 0.3kg object is:
p2_initial = m2 * v2 = 0
Therefore, the equation becomes:
p1_initial = p1_final + p2_final
Substituting the values into the equation:
m1 * v1 = m1 * v1' + m2 * v2'
Now, we can solve for the final velocity of the 0.3kg object (v2'):
v2' = (m1 * v1 - m1 * v1') / m2
Plugging in the given values:
v2' = (0.25kg * 4m/s - 0.25kg * 2m/s) / 0.3kg
v2' = (1kg*m/s - 0.5kg*m/s) / 0.3kg
v2' = 0.5kg*m/s / 0.3kg
v2' = 1.67m/s
The speed of the 0.3kg object after the collision is 1.67m/s (rounded to two decimal places).
b) To calculate the direction of motion for one of the objects after the collision, you need to analyze the given information about the initial and final velocities. In this case, the 0.25kg object was initially moving east (positive direction) at 4m/s and ended up moving east (positive direction) at 2m/s. Therefore, we can conclude that the 0.25kg object continued moving in the same direction after the collision.