Answer:
Velocity is 1.73 m/s along 54.65° south of east.
Explanation:
Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.
Here momentum is conserved.
Initial momentum = Final momentum
Initial momentum = m x 2i + m x (-1)i = m i
Final momentum = m x v + m x 1.41 j = mv + 1.41 m j
Comparing
mi = mv + 1.41 m j
v = i - 1.41 j
Magnitude of velocity
Direction,
Velocity is 1.73 m/s along 54.65° south of east.
Explanation:
It is given that,
Mass of golf club, m₁ = 210 g = 0.21 kg
Initial velocity of golf club, u₁ = 56 m/s
Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg
After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.
Initial momentum of golf ball,
After the collision, final momentum
Using the conservation of momentum as :
v = 63.91 m/s
So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.