Consider two blocks on a horizontal frictionless surface. Block A has a mass of
ID: 1366368 • Letter: C
Question
Consider two blocks on a horizontal frictionless surface. Block A has a mass of 70.0 kg and block B has a mass of 45.0 kg . Block A is at rest but block B is moving in a straight line at 14.0 m/s toward Block A and eventually collides with block A. After the impact, Block B is found to be traveling with a velocity of 6.00 m/s at an angle of 52.1 degrees with respect to its initial direction of motion. Consider air resistance to be negligible.
A) What is the magnitude of the velocity of Block A, 1.73 seconds after the collision?
Explanation / Answer
Since the block A and B are on the frictionless surface after the collision the block A moves with constant speed
then speed of block A after the collision is VA
Given that UA = 0 m/s
UB = 14 m/s
mA = 70 kg
mB = 45 kg
Apply law of conservation of momentum
Along X-axis
45*14 = (70*VA*cos(theta)) + (45*6*cos(52.1))
VA*cos(theta) = 6.63 m/s..................(1)
Along Y-axis
70*VA*sin(theta) = 70*6*sin(52.1)
VA*sin(theta) = 4.73 m/s..................(2)
(2)/(1) = tan(theta) = 4.73/6.63 =
theta = 35.53 degrees below theoriginal direction
VA = 8.13 m/s...is the answer