A 2.0-kg ball moving at 10 m/s makes an off-center collision with a 3.0-kg ball that is initially at rest. After the collision, the 2.0-kg ball is deflected at an angle of 30° from its original direction of motion and the 3.0-kg ball is moving at 4.0 m/s. Find the speed of the 2.0-kg ball and the direction of the 3.0-kg ball after the collision (angle from original direction of motion)
Apparently the answer is 5.3 at 26 degrees but I don't understand why?
m1=2 kg, v = 10 m/s, m2 = 3 kg, α =30º, u2 = 4 m/s. Find β and u1.
Take the x axis to be the original direction of motion of 2kkg ball. The equations expressing the conservation of momentum for the components in the x and y directions separately are
p(x): m1•v = m1•u1•cos α + m2•u2•cos β,
p(y): 0 = m1•u1•sin α - m2•u2•sin β.
cos β = (m1•v - m1•u1•cos α)/ m2•u2,
sin β = m1•u1•sin α/ m2•u2,
sin β/ cos β = tan β = m1•u1•sin α/ m1•v(1-cos α) =
=sin α/ (1-cos α) = 0.5/(1-0.866) = 3.73,
α = arctan 3.73 = 75º.
u1 = m2•u2•sin β / m1•sin α = 3•4•sin75º/2•sin30º = 11.6 m/s.
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.
Step 1: Calculate the initial momentum before the collision.
The formula for momentum is P = m * v, where P is momentum, m is mass, and v is velocity.
The momentum of the 2.0-kg ball before the collision is:
P1 = m1 * v1 = 2.0 kg * 10 m/s = 20 kg.m/s
The momentum of the 3.0-kg ball before the collision is:
P2 = m2 * v2 = 3.0 kg * 0 m/s = 0 kg.m/s (since it is initially at rest)
Step 2: Calculate the total momentum after the collision.
The total momentum after the collision should be equal to the total momentum before the collision, due to the conservation of momentum.
P_total = P1 + P2 = 20 kg.m/s + 0 kg.m/s
P_total = 20 kg.m/s
Step 3: Use the given information to find the speed of the 2.0-kg ball after the collision.
Let's assume the speed of the 2.0-kg ball after the collision is v1f, and the speed of the 3.0-kg ball after the collision is v2f.
Since the 2.0-kg ball is deflected at an angle of 30°, we can use trigonometry to calculate the horizontal and vertical components of its velocity.
The horizontal component of the velocity (v1fx) is v1f * cos(30°), where cos(30°) = √3/2.
So, v1fx = v1f * √3/2.
The vertical component of the velocity (v1fy) is v1f * sin(30°), where sin(30°) = 1/2.
So, v1fy = v1f * 1/2.
Step 4: Apply the conservation of kinetic energy to find v1f.
The formula for kinetic energy is KE = (1/2) * m * v^2, where KE is kinetic energy.
The kinetic energy before the collision is given by:
KE1 = (1/2) * m1 * v1^2
KE1 = (1/2) * 2.0 kg * (10 m/s)^2
KE1 = 100 J
The kinetic energy after the collision is given by:
KE1f = (1/2) * m1 * v1f^2
KE1f = (1/2) * 2.0 kg * v1f^2
Since kinetic energy is conserved, we have:
KE1 = KE1f
Substituting the values, we get:
100 J = (1/2) * 2.0 kg * v1f^2
100 J = v1f^2
v1f = √100 J = 10 m/s
Therefore, the speed of the 2.0-kg ball after the collision is 10 m/s.
Step 5: Use the given information to find the direction of the 3.0-kg ball after the collision.
Since the 3.0-kg ball is moving, we can calculate the angle it makes with its original direction of motion, using trigonometry.
Let's assume the angle is θ. We can use the equation:
tan(θ) = (vertical component of velocity) / (horizontal component of velocity)
Using the values we calculated earlier:
tan(θ) = (v1fy) / (v1fx)
tan(θ) = (v1f * 1/2) / (v1f * √3/2)
tan(θ) = 1/√3
θ = arctan(1/√3)
Using a calculator, we find θ ≈ 30°.
Therefore, the direction of the 3.0-kg ball after the collision is 30° from its original direction of motion.
To find the speed of the 2.0-kg ball after the collision, we can use the principle of conservation of linear momentum.
The principle of conservation of linear momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be written as:
(m1 * v1)initial + (m2 * v2)initial = (m1 * v1)final + (m2 * v2)final
Where m1 and m2 are the masses of the balls, v1 and v2 are their respective velocities before and after the collision.
In this case, the mass of the 2.0-kg ball (m1) is 2.0 kg, its initial velocity (v1) is 10 m/s, and the mass of the 3.0-kg ball (m2) is 3.0 kg. The 3.0-kg ball is initially at rest, so its initial velocity (v2) is 0 m/s.
Let's substitute these values into the conservation of linear momentum equation:
(2.0 kg * 10 m/s) + (3.0 kg * 0 m/s) = (2.0 kg * v1)final + (3.0 kg * 4.0 m/s)
20 kg·m/s = (2.0 kg * v1)final + 12 kg·m/s
Now, let's rearrange the equation to solve for (2.0 kg * v1)final:
(2.0 kg * v1)final = 20 kg·m/s - 12 kg·m/s
(2.0 kg * v1)final = 8 kg·m/s
Now we can solve for the speed of the 2.0-kg ball. Since the mass of the ball is 2.0 kg, we can divide both sides of the equation by 2.0 kg:
v1final = 8 kg·m/s / 2.0 kg
v1final = 4 m/s
Therefore, the speed of the 2.0-kg ball after the collision is 4 m/s.
Now, let's find the direction of the 3.0-kg ball after the collision. We are given that the 2.0-kg ball is deflected at an angle of 30° from its original direction of motion. This means that the direction of the 3.0-kg ball after the collision will be 180° - 30° = 150° from its original direction of motion.
Therefore, the direction of the 3.0-kg ball after the collision is 150° from its original direction of motion.