If I take a glob of mud and throw it against a wall with the speed of V and stops in time T and than I take a ball and throw it and it hits the wall with speed V and leaves the wall a time T later at a speed of 0.5V and they both have same value and mass, which one involves a greater total force on the object.
I think it is the ball but to be honest I don't understand why-I'm thinking because it is pushed off the wall once it hits it? Newton's laws?
Could you give me some direction, please?
force*time=changemomentum
if mv was the mud ball momentum
then mv+1/2 mv was the CHANGE of momentum for the ball.
Remember Vf for the ball= -V/2, its intial velocity was V
changemomentum=m (Vf-Vi)=m(3/2 V)
Thank you
looking for
To understand which object involves a greater total force, let's consider Newton's Second Law of Motion. According to this law, the force acting on an object is equal to the product of its mass and its acceleration. F = m * a.
In the case of the mud glob, when it hits the wall and stops, its acceleration changes from V/T to 0 m/s^2. This means that the change in velocity (Δv) is equal to -V in magnitude and occurs over a time period (Δt) equal to T. Using the formula for acceleration (a = Δv / Δt), we can calculate the acceleration of the mud glob: a = (-V) / T.
The force acting on the mud glob can be calculated by multiplying its mass (m) by its acceleration (a). Let's say the mass of the mud glob is M. Therefore, the force acting on the mud glob is F_mud = M * a = M * (-V) / T.
Now, let's analyze the ball scenario. When the ball hits the wall, it changes its velocity from V to 0.5V over a time period T. So, the change in velocity (Δv) is equal to (0.5V - V) = -0.5V, and it occurs in a time (Δt) equal to T. The acceleration (a) of the ball can be calculated as a = Δv / Δt = (-0.5V) / T.
Again, let's assume the mass of the ball is M. Therefore, the force acting on the ball is F_ball = M * a = M * (-0.5V) / T.
Comparing the two forces, F_mud and F_ball, we need to determine which one is greater. To do that, we can look at their ratios:
F_ball / F_mud = (M * (-0.5V) / T) / (M * (-V) / T)
= (-0.5V) / (-V)
= 0.5
Therefore, the ratio of forces is 0.5, which means that the force exerted on the ball is only half of the force exerted on the mud glob.
In conclusion, the ball involves a greater total force on the object compared to the mud glob when they hit the wall with the same speed and mass.