A 4.00 kg ball traveling North with a momentum of 11.2 kg.m/s collides with a 6.0 kg ball traveling
West with a momentum of 18 kg.m/s. The two balls stick together and head off as one. What is the
momentum of the combined masses? What is the velocity of the combined masses?
Well, let's calculate the momentum first. To find the total momentum, we can add up the individual momenta of the two balls before the collision.
The momentum of the first ball (4.00 kg) traveling North with a momentum of 11.2 kg·m/s is 4.00 kg * 11.2 kg·m/s = 44.8 kg·m/s.
The momentum of the second ball (6.0 kg) traveling West with a momentum of 18 kg·m/s is 6.0 kg * 18 kg·m/s = 108 kg·m/s.
Now, let's add these two momenta together to find the total momentum: 44.8 kg·m/s + 108 kg·m/s = 152.8 kg·m/s.
So, the momentum of the combined masses is 152.8 kg·m/s.
To find the velocity of the combined masses, we can use the law of conservation of momentum. Since the balls stick together after the collision, the total momentum before the collision must be equal to the total momentum after the collision.
The total mass of the combined masses is 4.00 kg + 6.0 kg = 10.00 kg.
Now, we divide the total momentum (152.8 kg·m/s) by the total mass (10.00 kg) to find the velocity:
152.8 kg·m/s / 10.00 kg = 15.28 m/s.
So, the velocity of the combined masses is 15.28 m/s.
And just remember, when two balls stick together, it's always a "balls-to-the-wall" moment!
To find the momentum of the combined masses, we can use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the momentum of the 4.00 kg ball as P1 and the momentum of the 6.0 kg ball as P2.
Given:
Mass of the 4.00 kg ball, m1 = 4.00 kg
Momentum of the 4.00 kg ball, P1 = 11.2 kg.m/s
Mass of the 6.0 kg ball, m2 = 6.0 kg
Momentum of the 6.0 kg ball, P2 = 18 kg.m/s
Before the collision:
Total momentum before the collision, P_total_before = P1 + P2
After the collision, the two balls stick together and move as one.
Let the mass of the combined system be M, and the velocity of the combined system be V.
After the collision:
Total momentum after the collision, P_total_after = M * V
According to the law of conservation of momentum, P_total_before = P_total_after.
P1 + P2 = M * V
Substituting the given values, we have:
11.2 kg.m/s + 18 kg.m/s = M * V
Now, we can solve for the momentum of the combined masses and the velocity of the combined masses.
To find the momentum of the combined masses, we need to first find the total mass of the two balls together and then calculate the momentum.
Step 1: Calculate the total mass
Since the two balls stick together after the collision, their masses will simply add up. So, the total mass (m) of the combined masses will be:
m = mass of ball 1 + mass of ball 2
= 4.00 kg + 6.0 kg
= 10.0 kg
Step 2: Calculate the momentum
The momentum (p) of the combined masses can be found by adding up the individual momenta of the balls before the collision. The momentum is a vector quantity, so we need to consider both magnitude and direction.
The momentum of ball 1 (p1) is given as 11.2 kg.m/s and it is traveling North. The momentum of ball 2 (p2) is 18 kg.m/s and it is traveling West. Since North and West are perpendicular directions, we can use the Pythagorean theorem to find the magnitude of the resultant momentum.
Magnitude of resultant momentum = sqrt( (p1)^2 + (p2)^2 )
= sqrt( (11.2 kg.m/s)^2 + (18 kg.m/s)^2 )
≈ 21.21 kg.m/s (rounded to two decimal places)
Step 3: Calculate the velocity
To find the velocity (v) of the combined masses, we divide the momentum by the total mass:
v = p/m
= (21.21 kg.m/s) / (10.0 kg)
≈ 2.12 m/s (rounded to two decimal places)
Therefore, the momentum of the combined masses is approximately 21.21 kg.m/s and the velocity of the combined masses is approximately 2.12 m/s.
M1*V1 + M2*V2 = M1*V + M2*V
11.2i - 18 = 4*V + 5*V = 9V
V = -2 + 1.24i = 2.36m/s[31.9o] N. of
W. = 2.36m/s[148.1o] CCW
Momentum = M1*V + M2*V =
4*2.36[31.9] + 6*2.36[31.9] =
9.44[31.9] + 14.2[31.9o]=23.6m/s[31.9o] N. of W.