Calculating Acceleration: Solving for the Force of a Clam

[Phoenixtears]

SOLVED

Homework Statement

Sea clams move by pushing water in one direction so that the water pushes them in the opposite direction. In doing so a 0.5 kg clam can accelerate 0.6 kg of water from rest to a speed of 1.9 m/s in 0.43 seconds. Calculate the magnitude of the clam's acceleration

Homework Equations

F= ma
Vf^2= V0^2 + 2ax
Vf= V0 + at
x= V0*t + .5a(t^2)

The Attempt at a Solution

I'm not sure why my answer isn't checking out. The initial velocity is 0. The final velocity is 1.9. The time is .43. These all fit into the third equation: 1.9= 0 + .43a. a then equals 4.42. This, however, is not working as my answer. What am I missing?

Thanks in advance!

~Phoenix

 

[ritwik06]

Homework Statement

Sea clams move by pushing water in one direction so that the water pushes them in the opposite direction. In doing so a 0.5 kg clam can accelerate 0.6 kg of water from rest to a speed of 1.9 m/s in 0.43 seconds. Calculate the magnitude of the clam's acceleration

Homework Equations

F= ma
Vf^2= V0^2 + 2ax
Vf= V0 + at
x= V0*t + .5a(t^2)

The Attempt at a Solution

I'm not sure why my answer isn't checking out. The initial velocity is 0. The final velocity is 1.9. The time is .43. These all fit into the third equation: 1.9= 0 + .43a. a then equals 4.42. This, however, is not working as my answer. What am I missing?

Thanks in advance!

~Phoenix

Apply conservation of momentum
0.6*1.9=0.5*v
v=2.28
$$a=\frac{\Delta v}{\Delta t}$$
a=2.28/0.43
a=5.3
Does this help u?

 

[Ygggdrasil]

The clam accelerates the water from 0 to 1.9 m/s in 0.43s. You need to calculate the acceleration of the clam, not the water. (hint: remember Newton's third law)

 

[Phoenixtears]

OH! Alrighty, gotcha. That makes sense to me. Thank you so much. I realize now that I was mixing variables that shouldn't be used to solve for this.

Thank you so very much!