A block of mass m = 3.6 kg, moving on a frictionless surface with a speed vi=9.3m/s makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 3.6 kg block recoils with a speed of vf=2.7m/s. In Figure 8.2, the speed of the block of mass M after the collision is closest to
7.6 m/s
To find the speed of the block of mass M after the collision, 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 speed of the block of mass M after the collision as Vf(M).
Before the collision:
The momentum of the block with mass m is given by p(m) = m * vi, where vi is the initial speed.
The momentum of the block with mass M is given by p(M) = 0 since it is at rest initially.
After the collision:
The momentum of the block with mass m is given by p(m) = m * vf, where vf is the final speed.
The momentum of the block with mass M is given by p(M) = M * Vf(M), where Vf(M) is the final speed.
Using the law of conservation of momentum, we can equate the total momentum before the collision to the total momentum after the collision:
m * vi + M * 0 = m * vf + M * Vf(M)
From the given information, we can substitute the values:
3.6 kg * 9.3 m/s + 0 = 3.6 kg * 2.7 m/s + M * Vf(M)
By solving this equation, we can find the value of Vf(M), which represents the speed of the block of mass M after the collision.
To find the speed of the block of mass M after the collision, we can use the principle of conservation of momentum and the concept of elastic collisions.
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
The momentum is given by the product of mass and velocity: momentum = mass x velocity.
Let's denote the mass of the block of mass M as M and its velocity after the collision as V.
Before the collision:
The momentum of the 3.6 kg block: momentum1 = m * vi (since it is moving)
The momentum of the block of mass M: momentum2 = 0 (since it is at rest)
After the collision:
The momentum of the 3.6 kg block: momentum1' = m * vf (since it is recoiling)
The momentum of the block of mass M: momentum2' = M * V
Using the conservation of momentum, we can equate the total momentum before and after the collision:
momentum1 + momentum2 = momentum1' + momentum2'
m * vi + 0 = m * vf + M * V
Substituting the given values: m = 3.6 kg, vi = 9.3 m/s, vf = 2.7 m/s, and solving for V:
3.6 kg * 9.3 m/s = 3.6 kg * 2.7 m/s + M * V
33.48 kg m/s = 9.72 kg m/s + M * V
M * V = 33.48 kg m/s - 9.72 kg m/s
M * V = 23.76 kg m/s
Since the collision is perfectly elastic, kinetic energy is conserved. Therefore, we can use the concept of conservation of kinetic energy:
The initial kinetic energy is given by: (1/2) * m * vi^2
The final kinetic energy is given by: (1/2) * m * vf^2 + (1/2) * M * V^2
Using the given values, we can substitute and solve for V:
(1/2) * 3.6 kg * (9.3 m/s)^2 = (1/2) * 3.6 kg * (2.7 m/s)^2 + (1/2) * M * V^2
121.212 kg m^2/s^2 = 32.868 kg m^2/s^2 + (1/2) * M * V^2
88.344 kg m^2/s^2 = (1/2) * M * V^2
Multiplying both sides of the equation by 2:
176.688 kg m^2/s^2 = M * V^2
From the equation obtained earlier: M * V = 23.76 kg m/s
Substituting M * V with 23.76 kg m/s, we get:
176.688 kg m^2/s^2 = (23.76 kg m/s) * V
Solving for V:
V = 176.688 kg m^2/s^2 / 23.76 kg m/s
V ≈ 7.43 m/s
Therefore, the speed of the block of mass M after the collision is closest to 7.43 m/s.