A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
1 answer:
To determine the speed 
of the combined blocks m1 and m2 post-collision, we will apply the principle of momentum conservation. Initially, only block 1 possesses momentum since block 2 is at rest, which can be expressed as:
After the collision, both blocks merge and are effectively one mass
moving at a speed :
Using momentum conservation:
We can rewrite this as:
This leads to:
As the blocks hit the rough terrain, they start to slow down due to friction
. The friction does work to bring the two blocks to a stop over a distance d:
where d represents the distance traveled before they come to a halt.
The total kinetic energy of the blocks just before they hit the rough surface is:
Once the blocks cease moving, the kinetic energy is lost as their speed drops to zero; according to the work-energy principle, the loss in kinetic energy equates to the work done by friction:
From this equation, we can derive the coefficient of kinetic friction:
After the collision, both blocks merge and are effectively one mass
Using momentum conservation:
We can rewrite this as:
This leads to:
As the blocks hit the rough terrain, they start to slow down due to friction
where d represents the distance traveled before they come to a halt.
The total kinetic energy of the blocks just before they hit the rough surface is:
Once the blocks cease moving, the kinetic energy is lost as their speed drops to zero; according to the work-energy principle, the loss in kinetic energy equates to the work done by friction:
From this equation, we can derive the coefficient of kinetic friction:
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a) The student's speed after jumping is 1.07 m/s
b) The final speed of the laser is 10.4 m/s
Explanation:
a)
This issue can be approached through the momentum conservation principle: In the absence of external forces, the combined momentum of the student and the laser must remain unchanged. Hence, we can express:
where:
The initial momentum is zero
m = 42 kg signifies the mass of the laser
v = 1.5 m/s is the laser's final velocity
M = 59 kg is the mass of the student
V denotes the student's final velocity
Solving this for V, we can determine the student's speed:
Thus, the student's final speed calculates to 1.07 m/s.
b)
Here, both the laser and the student have a combined speed of 3.1 m/s prior to the student's jump; thus, the initial momentum isn't zero.
<pSo, we formulate the equation of momentum conservation as:where:
m = 42 kg denotes the mass of the laser
M = 59 kg is the student’s mass
u = 3.1 m/s is their starting velocity
V = -2.1 m/s indicates the student's speed post-jump (she jumps backward)
v signifies the laser's final speed
When we resolve for v, we have:
Learn more about momentum:
Answer:
Part A) Electric fields at the designated point due to charges q₁ and q₂:
E₁ = 33.75 * 10³ N/C (-j), E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C
Part B) The overall electric field at P (Ep)
Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C
Explanation:
Conceptual analysis
The electric field at point P caused by a point charge is calculated as:
E = k*q/d²
E: Electric field measured in N/C
q: charge magnitude in Newtons (N)
k: electric constant measured in N*m²/C²
d: distance from the charge q to point P in meters (m)
Equivalence:
1 nC = 10⁻⁹ C
1 cm = 10⁻² m
Data:
k = 9 * 10⁹ N*m²/C²
q₁ = -6.00 nC = -6 * 10⁻⁹ C
q₂ = +3.00 nC = +3 * 10⁻⁹ C
d₁ = 4 cm = 4 * 10⁻² m
d₂ = 5 * 10⁻² m
Part A) Calculation for electric fields at point from q₁ and q₂:
Refer to the attached illustration:
E₁: Electric Field at point P(0,4) cm due to charge q₁. Since q₁ is negative (q₁-), the electric field approaches the charge.
E₂: Electric Field at point P(0,4) cm due to charge q₂. Since q₂ is positive (q₂+), the electric field emanates from the charge.
E₁ = k*q₁/d₁² = 9 * 10⁹ * 6 * 10⁻⁹ / (4 * 10⁻²)² = 33.75 * 10³ N/C
E₂ = k*q₂/d₂²= 9 * 10⁹ * 3 * 10⁻⁹ / (5 * 10⁻²)² = 10.8 * 10³ N/C
E₁ = 33.75 * 10³ N/C (-j)
E₂x = E₂cosβ = 10.8 * (3/5) = 6.48 * 10³ N/C
E₂y = E₂sinβ = 10.8 * (4/5) = 8.64 * 10³ N/C
E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C
Part B) Calculation for net electric field at P (Ep)
The electric field at point P from multiple point charges is the vector sum of the individual electric fields.
Ep = Epx (i) + Epy (j)
Epx = E₂x = 6.48 * 10³ N/C (-i)
Epy = E₁y + E₂y = (33.75 * 10³ (-j) + 8.64 * 10³ (+j)) N/C = 25.11 * 10³ (-j) N/C
Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C
Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C
