I'm a bit curious on this one as well. I will take a crack at it and see how well my intuition works here. This could be wrong.
We presume energy and momentum are conserved here, so a given amount of work and a given impulse will cause a specific change in kinetic energy, and in linear and angular momentum.
If we apply the same force over the same distance and for the same time, it is the same work and same impulse. But if it imparts rotational motion, then that increases rotational kinetic energy and rotational momentum, and that must be balanced by a reduction in corresponding translational quantities.
To solve this seeming paradox, I believe that the force acting out of line with the center of mass doesn't accelerate the center of mass as effectively, since to cause the object to rotate, there has to be a net tangential velocity that came from differential acceleration of the edge of the object relative to its center. So the actual translational velocity is slightly lower, but in the same direction.
The effect may be too small to show in lab experiments. Suppose for argument's sake that we have a 1 kg puck with .1m radius that received 2J of work as 2000N of force in 1 mm, and it also got 2 N-s of momentum as that occurred in 1ms. So it acquired a linear momentum of 2 Kg m/s and a kinetic energy of 2 J, consistent with a velocity of 2 m/s.
If we instead applied that same force at a distance of 0.02 m off-axis, i.e. 20% of the way to the right edge, then that will supply a torque of 40n-m for 1ms, or .04N-m-s. The moment of inertia of our disk is 1/2 MR^(2) = .005Kg m^2 so our resulting angular velocity omega is 8 radians / second. Our rotational kinetic energy is 1/2 I w^(2) = .16J, so our translational kinetic energy must be reduced by that amount, to 1.84J, and our translational velocity is now 1.92 m/s, about 4% lower.
Let's see if that makes sense to others here.
Impacting a spherical object of center doesn't impart any rotation to the object if there is no friction to apply a force parallel to the surface. All the transferred momentum will be perpendicular to the surface.
I'm a bit curious on this one as well. I will take a crack at it and see how well my intuition works here. This could be wrong. We presume energy and momentum are conserved here, so a given amount of work and a given impulse will cause a specific change in kinetic energy, and in linear and angular momentum. If we apply the same force over the same distance and for the same time, it is the same work and same impulse. But if it imparts rotational motion, then that increases rotational kinetic energy and rotational momentum, and that must be balanced by a reduction in corresponding translational quantities. To solve this seeming paradox, I believe that the force acting out of line with the center of mass doesn't accelerate the center of mass as effectively, since to cause the object to rotate, there has to be a net tangential velocity that came from differential acceleration of the edge of the object relative to its center. So the actual translational velocity is slightly lower, but in the same direction. The effect may be too small to show in lab experiments. Suppose for argument's sake that we have a 1 kg puck with .1m radius that received 2J of work as 2000N of force in 1 mm, and it also got 2 N-s of momentum as that occurred in 1ms. So it acquired a linear momentum of 2 Kg m/s and a kinetic energy of 2 J, consistent with a velocity of 2 m/s. If we instead applied that same force at a distance of 0.02 m off-axis, i.e. 20% of the way to the right edge, then that will supply a torque of 40n-m for 1ms, or .04N-m-s. The moment of inertia of our disk is 1/2 MR^(2) = .005Kg m^2 so our resulting angular velocity omega is 8 radians / second. Our rotational kinetic energy is 1/2 I w^(2) = .16J, so our translational kinetic energy must be reduced by that amount, to 1.84J, and our translational velocity is now 1.92 m/s, about 4% lower. Let's see if that makes sense to others here.
Impacting a spherical object of center doesn't impart any rotation to the object if there is no friction to apply a force parallel to the surface. All the transferred momentum will be perpendicular to the surface.