How to create a mechanical equivalent to dropping object onto steel floor

(I was informed on the Physics board that this is an Engineering question, therefore they closed it there, and I moved it here)

First, I have next to zero formal physics training… I opted for all math & electronics electives back in my 1970s high school curriculum.

Goal:
My goal is to create a replacement for a "vertical drop test" by accelerating an impact mass in the horizontal plane.

The existing Vertical Drop Test machine that I want to replace drops a small product 1 meter onto a steel plate, but the orientation of the product is not predictable, and it needs to be.

By creating a Horizontal impact machine to substitute for the vertical drop, the product can then be pre-positioned on the machine so that the impact happens at a precise location on the product being tested.

My main challenge is understanding how to calculate what mass to accelerate at the "product", and at what velocity.

My example follows:
Assuming the object being dropped vertically is a 100g solid plastic ball. I calculate the impact speed of a 1 meter vertical drop to be 4.429 m/s, resulting in 0.98 J kinetic energy.

On my proposed horizontal impact machine, I would have the same ball resting on a horizontal shelf (without constraint), and impact it horizontally with a given mass/velocity. The "ball" will then move away from the point of impact, just like it would bounce when dropped on the floor. If that horizontal impact mass is 250g (with essentially no deflection) moving at 2.8 m/s (which is still 0.98 J Kinetic Energy, I believe), are the impact forces equivalent to the vertical drop regardless of the "give" (deflection) the "plastic ball" may have?

I know that if the "ball" were stationary, and I moved an infinite mass (the floor) up towards the
ball" at the same speed as the ball being dropped vertically, the impact would be the same. (right?)

If I use a lower impact mass, I would need a higher velocity to equate to the same impact Force, is that correct?

Does the impact mass need to be many times (10x or greater) the mass of the product being impacted, in order to cause the equivalent impact Force?

I love to learn, but again, my physics background is minimal… I’d appreciate any lay-language help, with formulas (or even a link to a web site with a calculator for such a problem).

/edit/

A bit more "detail". This is for testing watch buttons… it’s called a "Button Focused Drop Test", and is not an ISO regulated test.

The problem with the "normal" way of performing this test, is that the machine releases the "sample" as it’s falling, but well before it impacts the target… which in this case is a steel plate. The watch almost always tends to "tumble", making the impact point/position unpredictable. In addition, when watches are "square", and have multiple buttons on a flat side are tested, it’s very hard to isolate a specific button. Also… once it’s dropped and impacts the steel plate, it bounces elsewhere, possibly incurring additional damage not related to the initial impact.

If the watch is laid on a horizontal "shelf", and the impact device is pre-aligned so that the impact position is precisely controlled, and once the impact occurs it falls into a padded receptacle, the initial impact is all that happens… and the results are extremely predictable.