Recent content by skoczek77

yes, you are right, 2 hours ago I thought there was not enough data to solve it; thank you very much for help have a nice day ;)

I've already done it, tell me if it's good: 0,5·m·v1²=0,5·(M+m)·v2² m·v1²=(M+m)·v2² v2=(m·v1)/(M-m) m·v1²=(M+m)·(m·v1)²/(M-m)² m·v1²=(M+m)·m²·v1²/(M-m)² 1=(M+m)·m/(M-m)² (M-m)²=m·M+m² M²-2·M·m+m²=m·M+m² M²-2·M·m=m·M M-2·m=m M=3·m so m/M=1/3

with kinetic energy but idk how

p=mv1 before and p=-mv2 +Mv2 after

i dont know how