A mass of 100g strikes the wall with speed 5m/s at an angle of 60 and rebounds with the same speed. If the contact time is 2*10^-3 sec., what is the force applied by the wall?
To find the force applied by the wall, we can use Newton's second law of motion, which states that force (F) is equal to the rate of change of momentum.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v). We can calculate the initial momentum (p1) and the final momentum (p2) of the mass.
Given:
Mass (m) = 100g = 0.1 kg (convert grams to kilograms)
Initial speed (v1) = 5 m/s
Angle (θ) = 60°
Contact time (Δt) = 2 * 10^-3 sec
Step 1: Calculate the initial momentum (p1)
p1 = m * v1
Step 2: Calculate the final momentum (p2)
Since the mass rebounds with the same speed, the magnitude of the final momentum is the same as the initial momentum.
p2 = p1
Step 3: Calculate the change in momentum (Δp)
Δp = p2 - p1
Step 4: Calculate the force (F)
The force applied by the wall can be calculated using the formula:
F = Δp / Δt
Now let's go through the calculation:
Step 1: Calculate the initial momentum (p1)
p1 = m * v1
= 0.1 kg * 5 m/s
= 0.5 kg·m/s
Step 2: Calculate the final momentum (p2)
Since the mass rebounds with the same speed, the magnitude of the final momentum is the same as the initial momentum.
p2 = p1
= 0.5 kg·m/s
Step 3: Calculate the change in momentum (Δp)
Δp = p2 - p1
= 0.5 kg·m/s - 0.5 kg·m/s
= 0 kg·m/s
Step 4: Calculate the force (F)
F = Δp / Δt
= 0 kg·m/s / (2 * 10^-3 sec)
= 0 kg·m/s / 0.002 sec
= 0 N
Hence, the force applied by the wall is 0 Newtons.
Let θ=60º be the angle that the velocity makes with the wall.
Δp(x)= p2(x)-p1(x)= m•v(x) – (-m•v(x)) =2m•v(x)= 2m•v•sin θ
Δp(y)= p2(y)-p1(y)=0
F•Δt=Δ p(x)= 2m• v•sin θ
F=2m• v•sin θ/Δt